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Every infinite set has a denumerable subset. In particular, \( {\aleph }_{0} \leq \alpha \) for every infinite cardinal number \( \alpha \) . | If \( B \) is a finite subset of the infinite set \( A \), then \( A - B \) is nonempty. For each finite subset \( B \) of \( A \), choose an element \( {x}_{B}{\varepsilon A} - B \) (Axiom of Choice). Let \( F \) be the set of all finite subsets of \( A \) and define a map \( f : F \rightarrow F \) by \( f\left( B\rig... | Yes |
Lemma 8.9. If \( \mathrm{A} \) is an infinite set and \( \mathrm{F} \) a finite set then \( \left| {\mathrm{A} \cup \mathrm{F}}\right| = \left| \mathrm{A}\right| \) . In particular, \( \alpha + \mathrm{n} = \alpha \) for every infinite cardinal number \( \alpha \) and every natural number (finite cardinal) \( \mathrm{n... | SKETCH OF PROOF. It suffices to assume \( A \cap F = \varnothing \) (replace \( F \) by \( F - A \) if necessary). If \( F = \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{n}}\right\} \) and \( D = \left\{ {{x}_{i} \mid i \in {\mathbf{N}}^{ * }}\right\} \) is a denumerable subset of \( A \) (Theorem 8.8), verify that \( f : A \... | No |
Theorem 8.10. If \( \alpha \) and \( \beta \) are cardinal numbers such that \( \beta \leq \alpha \) and \( \alpha \) is infinite, then \( \alpha + \beta = \alpha \) . | SKETCH OF PROOF. It suffices to prove \( \alpha + \alpha = \alpha \) (simply verify that \( \alpha \leq \alpha + \beta \leq \alpha + \alpha = \alpha \) and apply the Schroeder-Bernstein Theorem to conclude \( \alpha + \beta = \alpha ) \) . Let \( A \) be a set with \( \left| A\right| = \alpha \) and let \( \mathfrak{F}... | No |
Theorem 8.11. If \( \alpha \) and \( \beta \) are cardinal numbers such that \( 0 \neq \beta \leq \alpha \) and \( \alpha \) is infinite, then \( {\alpha \beta } = \alpha \) ; in particular, \( \alpha {\aleph }_{0} = \alpha \) and if \( \beta \) is finite \( {\aleph }_{0}\beta = {\aleph }_{0} \) . | SKETCH OF PROOF. Since \( \alpha \leq {\alpha \beta } \leq {\alpha \alpha } \) it suffices (as in the proof of Theorem 8.10) to prove \( {\alpha \alpha } = \alpha \) . Let \( A \) be an infinite set with \( \left| A\right| = \alpha \) and let \( \mathcal{F} \) be the set of all bijections \( f : X \times X \rightarrow ... | No |
Theorem 8.12. Let \( \mathrm{A} \) be a set and for each integer \( \mathrm{n} \geq 1 \) let \( {\mathrm{A}}^{\mathrm{n}} = \mathrm{A} \times \mathrm{A} \times \cdots \times \mathrm{A} \) (n factors).\n\n(i) If \( \mathrm{A} \) is finite, then \( \left| {\mathrm{A}}^{\mathrm{n}}\right| = {\left| \mathrm{A}\right| }^{\m... | SKETCH OF PROOF. (i) is trivial if \( \left| A\right| \) is finite and may be proved by induction on \( n \) if \( \left| A\right| \) is infinite (the case \( n = 2 \) is given by Theorem 8.11). | No |
Corollary 8.13. If \( \mathrm{A} \) is an infinite set and \( \mathrm{F}\left( \mathrm{A}\right) \) the set of all finite subsets of \( \mathrm{A} \), then \( \left| {F\left( A\right) }\right| = \left| A\right| \) . | PROOF. The map \( A \rightarrow F\left( A\right) \) given by \( a \mapsto \{ a\} \) is injective so that \( \left| A\right| \leq \left| {F\left( A\right) }\right| \) . For each \( n \) -element subset \( S \) of \( A \), choose \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \varepsilon {A}^{n} \) such that \( S = \left\{ {... | Yes |
Theorem 1.2. If \( \mathrm{G} \) is a monoid, then the identity element \( \mathrm{e} \) is unique. | SKETCH OF PROOF. If \( {e}^{\prime } \) is also a two-sided identity, then \( e = e{e}^{\prime } = {e}^{\prime } \). | No |
Proposition 1.3. Let \( \mathrm{G} \) be a semigroup. Then \( \mathrm{G} \) is a group if and only if the following conditions hold:\n\n(i) there exists an element \( \mathrm{e} \in \mathrm{G} \) such that \( \mathrm{{ea}} = \mathrm{a} \) for all \( \mathrm{a} \in \mathrm{G} \) (left identity element);\n\n(ii) for each... | SKETCH OF PROOF OF 1.3. ( \( \Rightarrow \) ) Trivial. ( \( \Leftarrow \) ) Note that Theorem 1.2(i) is true under these hypotheses. \( G \neq \varnothing \) since \( e \in G \) . If \( a \in G \), then by (ii) \( \left( {a{a}^{-1}}\right) \left( {a{a}^{-1}}\right) \) \( = a\left( {{a}^{-1}a}\right) {a}^{-1} = a\left( ... | Yes |
Proposition 1.4. Let \( \mathrm{G} \) be a semigroup. Then \( \mathrm{G} \) is a group if and only if for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{G} \) the equations \( \mathrm{{ax}} = \mathrm{b} \) and \( \mathrm{{ya}} = \mathrm{b} \) have solutions in \( \mathrm{G} \) . | PROOF. Exercise; use Proposition 1.3. | No |
Theorem 1.5. Let \( \mathrm{R}\left( \sim \right) \) be an equivalence relation on a monoid \( \mathrm{G} \) such that \( {\mathrm{a}}_{1} \sim {\mathrm{a}}_{2} \) and \( {\mathrm{b}}_{1} \sim {\mathrm{b}}_{2} \) imply \( {\mathrm{a}}_{1}{\mathrm{\;b}}_{1} \sim {\mathrm{a}}_{2}{\mathrm{\;b}}_{2} \) for all \( {\mathrm{... | PROOF OF 1.5. If \( {\bar{a}}_{1} = {\bar{a}}_{2} \) and \( {\bar{b}}_{1} = {\bar{b}}_{2}\left( {{a}_{i},{b}_{i}{\varepsilon G}}\right) \), then \( {a}_{1} \sim {a}_{2} \) and \( {b}_{1} \sim {b}_{2} \) by (20) of Introduction, Section 4. Then by hypothesis \( {a}_{1}{b}_{1} \sim {a}_{2}{b}_{2} \) so that \( \overline{... | Yes |
Theorem 1.6. (Generalized Associative Law) If \( \mathrm{G} \) is a semigroup and \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}}\varepsilon \mathrm{G} \) , then any two meaningful products of \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}} \) in this order are equal. | PROOF. We use induction to show that for every \( n \) any meaningful product \( {a}_{1}\cdots {a}_{n} \) is equal to the standard \( n \) product \( \mathop{\prod }\limits_{{i = 1}}^{n}{a}_{i} \) . This is certainly true for \( n = 1,2 \) . If \( n > 2 \), then by definition \( \left( {{a}_{1}\cdots {a}_{n}}\right) = ... | Yes |
Corollary 1.7. (Generalized Commutative Law) If \( \mathrm{G} \) is a commutative semigroup and \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}}\varepsilon \mathrm{G} \), then for any permutation \( {\mathrm{i}}_{1},\ldots ,{\mathrm{i}}_{\mathrm{n}} \) of \( 1,2,\ldots \mathrm{n},{\mathrm{a}}_{1}{\mathrm{a}}_{2}\c... | PROOF. Exercise. | No |
Theorem 1.9. If \( \mathrm{G} \) is a group [resp. semigroup, monoid] and \( \mathrm{a}\varepsilon \mathrm{G} \), then for all \( \mathrm{m},\mathrm{n}\varepsilon \mathbf{Z} \) [resp. \( {\mathrm{N}}^{ * },\mathrm{N} \) ]:\n\n(i) \( {\mathrm{a}}^{\mathrm{m}}{\mathrm{a}}^{\mathrm{n}} = {\mathrm{a}}^{\mathrm{m} + \mathrm... | SKETCH OF PROOF. Verify that \( {\left( {a}^{n}\right) }^{-1} = {\left( {a}^{-1}\right) }^{n} \) for all \( {n\varepsilon }\mathbf{N} \) and that \( {a}^{-n} = {\left( {a}^{-1}\right) }^{n} \) for all \( {n\varepsilon }\mathbf{Z} \) . (i) is true for \( m > 0 \) and \( n > 0 \) since the product of a standard \( n \) p... | No |
Theorem 2.3. Let \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) be a homomorphism of groups. Then\n\n(i) \( \mathrm{f} \) is a monomorphism if and only if \( \operatorname{Ker}\mathrm{f} = \{ \mathrm{e}\} \) ;\n\n(ii) \( \mathrm{f} \) is an isomorphism if and only if there is a homomorphism \( {\mathrm{f}}^{-1} :... | PROOF. (i) If \( f \) is a monomorphism and \( a \in \operatorname{Ker}f \), then \( f\left( a\right) = {e}_{H} = f\left( e\right) \) , whence \( a = e \) and \( \operatorname{Ker}f = \{ e\} \) . If \( \operatorname{Ker}f = \{ e\} \) and \( f\left( a\right) = f\left( b\right) \), then \( {e}_{H} = f\left( a\right) f{\l... | Yes |
Theorem 2.5. Let \( \mathrm{H} \) be a nonempty subset of a group \( \mathrm{G} \) . Then \( \mathrm{H} \) is a subgroup of \( \mathrm{G} \) if and only if \( {\mathrm{{ab}}}^{-1}\varepsilon \mathrm{H} \) for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{H} \) . | PROOF. ( \( \Leftarrow \) ) There exists \( {a\varepsilon H} \) and hence \( e = a{a}^{-1}{\varepsilon H} \) . Thus for any \( {b\varepsilon H},{b}^{-1} \) \( = e{b}^{-1}{\varepsilon H} \) . If \( a,{b\varepsilon H} \), then \( {b}^{-1}{\varepsilon H} \) and hence \( {ab} = a{\left( {b}^{-1}\right) }^{-1}{\varepsilon H... | Yes |
Corollary 2.6. If \( \mathrm{G} \) is a group and \( \left\{ {{\mathrm{H}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) is a nonempty family of subgroups, then \( \bigcap {\mathrm{H}}_{\mathrm{i}} \) is a subgroup of \( \mathrm{G} \) . \( {ieI} \) | PROOF. Exercise. | No |
Theorem 2.8. If \( \\mathrm{G} \) is a group and \( \\mathrm{X} \) is a nonempty subset of \( \\mathrm{G} \), then the subgroup \( \\langle \\mathrm{X}\\rangle \) generated by \( \\mathrm{X} \) consists of all finite products \( {\\mathrm{a}}_{1}{}^{{\\mathrm{n}}_{1}}{\\mathrm{a}}_{2}{}^{{\\mathrm{n}}_{2}}\\cdots {\\ma... | SKETCH OF PROOF. Show that the set \( H \) of all such products is a subgroup of \( G \) that contains \( X \) and is contained in every subgroup containing \( X \). Therefore \( H < \\langle X\\rangle < H. | No |
Theorem 3.1. Every subgroup \( \mathrm{H} \) of the additive group \( \mathbf{Z} \) is cyclic. Either \( \mathrm{H} = \langle 0\rangle \) or \( \mathrm{H} = \langle \mathrm{m}\rangle \), where \( \mathrm{m} \) is the least positive integer in \( \mathrm{H} \). If \( \mathrm{H} \neq \langle 0\rangle \), then \( \mathrm{... | PROOF. Either \( H = \langle 0\rangle \) or \( H \) contains a least positive integer \( m \). Clearly \( \langle m\rangle = \{ {km} \mid k \in \mathbf{Z}\} \subset H \). Conversely if \( h \in H \), then \( h = {qm} + r \) with \( q, r \in \mathbf{Z} \) and \( 0 \leq r < m \) (division algorithm). Since \( r = h - {qm... | Yes |
Every infinite cyclic group is isomorphic to the additive group \( \mathbf{Z} \) and every finite cyclic group of order \( \mathbf{r} \) is isomorphic to the additive group \( {\mathbf{Z}}_{\mathrm{m}} \) . | If \( G = \langle a\rangle \) is a cyclic group then the map \( \alpha : \mathbf{Z} \rightarrow G \) given by \( k \mapsto {a}^{k} \) is an epimorphism by Theorems 1.9 and 2.8. If Ker \( \alpha = 0 \), then \( \mathbf{Z} \cong G \) by Theorem 2.3 (i). Otherwise Ker \( \alpha \) is a nontrivial subgroup of \( \mathbf{Z}... | No |
Theorem 3.4. Let \( \mathrm{G} \) be a group and \( \mathrm{a}\varepsilon \mathrm{G} \) . If \( \mathrm{a} \) has infinite order, then\n\n(i) \( {\mathrm{a}}^{\mathrm{k}} = \mathrm{e} \) if and only if \( \mathrm{k} = 0 \) ;\n\n(ii) the elements \( {\mathrm{a}}^{\mathrm{k}}\left( {\mathrm{k}\varepsilon \mathbf{Z}}\righ... | SKETCH OF PROOF. (i)-(vi) are immediate consequences of the proof of Theorem 3.2. (vii) \( {\left( {a}^{k}\right) }^{m/k} = {a}^{m} = e \) and \( {\left( {a}^{k}\right) }^{r} \neq e \) for all \( 0 < r < m/k \) since otherwise \( {a}^{kr} = e \) with \( {kr} < k\left( {m/k}\right) = m \) contradicting (iii). Therefore,... | Yes |
Theorem 3.5. Every homomorphic image and every subgroup of a cyclic group \( \mathbf{G} \) is cyclic. In particular, if \( \mathrm{H} \) is a nontrivial subgroup of \( \mathrm{G} = \langle \mathrm{a}\rangle \) and \( \mathrm{m} \) is the least positive integer such that \( {\mathrm{a}}^{\mathrm{m}}\varepsilon \mathrm{H... | SKETCH OF PROOF. If \( f : G \rightarrow K \) is a homomorphism of groups, then \( \operatorname{Im}f = \langle f\left( a\right) \rangle \) . To prove the second statement simply translate the proof of Theorem 3.1 into multiplicative notation (that is, replace every \( t \in \mathbf{Z} \) by \( {a}^{t} \) throughout). ... | No |
Theorem 3.6. Let \( \mathrm{G} = \langle \mathrm{a}\rangle \) be a cyclic group. If \( \mathrm{G} \) is infinite, then \( \mathrm{a} \) and \( {\mathrm{a}}^{-1} \) are the only generators of \( \mathrm{G} \) . If \( \mathrm{G} \) is finite of order \( \mathrm{m} \), then \( {\mathrm{a}}^{\mathrm{k}} \) is a generator o... | SKETCH OF PROOF. It suffices to assume either that \( G = \mathbf{Z} \), in which case the conclusion is easy to prove, or that \( G = {Z}_{m} \) . If \( \left( {k, m}\right) = 1 \), there are \( c, d \in \mathbf{Z} \) such that \( {ck} + {dm} = 1 \) ; use this fact to show that \( \bar{k} \) generates \( {Z}_{m} \) . ... | No |
Theorem 4.2. Let \( \mathrm{H} \) be a subgroup of a group \( \mathrm{G} \) . (i) Right [resp. left] congruence modulo \( \mathrm{H} \) is an equivalence relation on \( \mathrm{G} \) . (ii) The equivalence class of \( \mathrm{a}\varepsilon \mathrm{G} \) under right [resp. left] congruence modulo \( \mathrm{H} \) is the... | PROOF OF 4.2. We write \( a \equiv b \) for \( a{ \equiv }_{r}b\left( {\;\operatorname{mod}\;H}\right) \) and prove the theorem for right congruence and right cosets. Analogous arguments apply to left congruence. (i) Let \( a, b, c \in G \) . Then \( a \equiv a \) since \( a{a}^{-1} = e \in H \) ; hence \( \equiv \) is... | Yes |
Theorem 4.5. If \( \mathrm{K},\mathrm{H},\mathrm{G} \) are groups with \( \mathrm{K} < \mathrm{H} < \mathrm{G} \), then \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack = \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \left\lbrack {\mathrm{H} : \mathrm{K}}\right\rbrack \) . If any two of these indices are fi... | PROOF. By Corollary \( {4.3G} = \mathop{\bigcup }\limits_{{i\varepsilon I}}H{a}_{i} \) with \( {a}_{i}{\varepsilon G},\left| I\right| = \left\lbrack {G : H}\right\rbrack \) and the cosets \( H{a}_{i} \) mutually disjoint (that is, \( H{a}_{i} = H{a}_{j} \Leftrightarrow i = j \) ). Similarly \( H = \mathop{\bigcup }\lim... | Yes |
Corollary 4.6. (Lagrange). If \( \mathrm{H} \) is a subgroup of a group \( \mathrm{G} \), then \( \left| \mathrm{G}\right| = \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \left| \mathrm{H}\right| \) . In particular if \( \mathrm{G} \) is finite, the order \( \left| \mathrm{a}\right| \) of \( \mathrm{a}\varepsilon... | PROOF. Apply the theorem with \( K = \langle e\rangle \) for the first statement. The second is a special case of the first with \( H = \langle a\rangle \) . | No |
Theorem 4.7. Let \( \mathrm{H} \) and \( \mathrm{K} \) be finite subgroups of a group \( \mathrm{G} \) . Then \( \left| \mathrm{{HK}}\right| = \) \( \left| \mathrm{H}\right| \left| \mathrm{K}\right| /\left| {\mathrm{H} \cap \mathrm{K}}\right| \) . | SKETCH OF PROOF. \( C = H \cap K \) is a subgroup of \( K \) of index \( n = \) \( \left| K\right| /\left| {H \cap K}\right| \) and \( K \) is the disjoint union of right cosets \( C{k}_{1} \cup C{k}_{2} \cup \cdots \cup C{k}_{n} \) for some \( {k}_{i} \in K \) . Since \( {HC} = H \), this implies that \( {HK} \) is th... | No |
Proposition 4.8. If \( \mathrm{H} \) and \( \mathrm{K} \) are subgroups of a group \( \mathrm{G} \), then \( \left\lbrack {\mathrm{H} : \mathrm{H} \cap \mathrm{K}}\right\rbrack \leq \) \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack \) . If \( \left\lbrack {\mathrm{G} : \mathrm{K}}\right\rbrack \) is finite, the... | SKETCH OF PROOF. Let \( A \) be the set of all right cosets of \( H \cap K \) in \( H \) and \( B \) the set of all right cosets of \( K \) in \( G \) . The map \( \varphi : A \rightarrow B \) given by \( \left( {H \cap K}\right) h \mapsto {Kh} \) \( \left( {h \in H}\right) \) is well defined since \( \left( {H \cap K}... | No |
Proposition 4.9. Let \( \\mathrm{H} \) and \( \\mathrm{K} \) be subgroups of finite index of a group \( \\mathrm{G} \) . Then \( \\left\\lbrack {\\mathrm{G} : \\mathrm{H} \\cap \\mathrm{K}}\\right\\rbrack \) is finite and \( \\left\\lbrack {\\mathrm{G} : \\mathrm{H} \\cap \\mathrm{K}}\\right\\rbrack \\leq \\left\\lbrac... | PROOF. Exercise; use Theorem 4.5 and Proposition 4.8. | No |
Theorem 5.1. If \( \mathrm{N} \) is a subgroup of a group \( \mathrm{G} \), then the following conditions are equivalent.\n\n(i) Left and right congruence modulo \( \mathbf{N} \) coincide (that is, define the same equivalence relation on \( \mathrm{G} \) );\n\n(ii) every left coset of \( \mathbf{N} \) in \( \mathbf{G} ... | PROOF. (i) \( \Leftrightarrow \) (iii) Two equivalence relations \( R \) and \( S \) are identical if and only if the equivalence class of each element under \( R \) is equal to its equivalence class under \( S \) . In this case the equivalence classes are the left and right cosets respectively of \( N \) . (ii) \( \Ri... | Yes |
Theorem 5.3. Let \( \mathrm{K} \) and \( \mathrm{N} \) be subgroups of a group \( \mathrm{G} \) with \( \mathrm{N} \) normal in \( \mathrm{G} \) . Then\n\n(i) \( \mathrm{N} \cap \mathrm{K} \) is a normal subgroup of \( \mathrm{K} \) ;\n\n(ii) \( \mathrm{N} \) is a normal subgroup of \( \mathrm{N} \vee \mathrm{K} \) ;\n... | PROOF. (i) If \( n \in N \cap K \) and \( a \in K \), then \( {an}{a}^{-1} \in N \) since \( N \vartriangleleft G \) and \( {an}{a}^{-1} \in K \) since \( K < G \) . Thus \( a\left( {N \cap K}\right) {a}^{-1} \subset N \cap K \) and \( N \cap K \vartriangleleft K \) . (ii) is trivial since \( N < N \vee K \) . (iii) Cl... | Yes |
Theorem 5.4. If \( \mathrm{N} \) is a normal subgroup of a group \( \mathrm{G} \) and \( \mathrm{G}/\mathrm{N} \) is the set of all (left) cosets of \( \mathrm{N} \) in \( \mathrm{G} \), then \( \mathrm{G}/\mathrm{N} \) is a group of order \( \left\lbrack {\mathrm{G} : \mathrm{N}}\right\rbrack \) under the binary opera... | PROOF. Since the coset \( {aN} \) [resp. \( {bN},{abN} \) ] is simply the equivalence class of \( a \in G \) [resp. \( b \in G,{ab} \in G \) ] under the equivalence relation of congruence modulo \( N \), it suffices by Theorem 1.5 to show that congruence modulo \( N \) is a congruence relation, that is, that \( {a}_{1}... | Yes |
Theorem 5.5. If \( f : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, then the kernel of \( \mathrm{f} \) is a normal subgroup of \( \mathrm{G} \) . Conversely, if \( \mathrm{N} \) is a normal subgroup of \( \mathrm{G} \), then the map \( \pi : \mathrm{G} \rightarrow \mathrm{G}/\mathrm{N} \) given by... | PROOF. If \( x \in \operatorname{Ker}f \) and \( a \in G \), then\n\n\[ f\left( {{ax}{a}^{-1}}\right) = f\left( a\right) f\left( x\right) f\left( {a}^{-1}\right) = f\left( a\right) {ef}{\left( a\right) }^{-1} = e \]\n\nand \( {ax}{a}^{-1}\varepsilon \operatorname{Ker}f \) . Therefore \( a\left( {\operatorname{Ker}f}\ri... | Yes |
Theorem 5.6. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups and \( \mathrm{N} \) is a normal subgroup of \( \mathrm{G} \) contained in the kernel of \( \mathrm{f} \), then there is a unique homomorphism \( \bar{\mathrm{f}} : \mathrm{G}/\mathrm{N} \rightarrow \mathrm{H} \) such that ... | PROOF OF 5.6. If \( {b\varepsilon aN} \), then \( b = {an},{n\varepsilon N} \), and \( f\left( b\right) = f\left( {an}\right) = f\left( a\right) f\left( n\right) \) \( = f\left( a\right) e = f\left( a\right) \), since \( N < \operatorname{Ker}f \) . Therefore, \( f \) has the same effect on every element of the coset \... | Yes |
Corollary 5.7. (First Isomorphism Theorem) If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, then \( \mathrm{f} \) induces an isomorphism \( \mathrm{G}/\operatorname{Ker}\mathrm{f} \cong \operatorname{Im}\mathrm{f} \) . | PROOF. \( f : G \rightarrow \operatorname{Im}f \) is an epimorphism. Apply Theorem 5.6 with \( N = \operatorname{Ker}f \) . \( \blacksquare \) | No |
Corollary 5.8. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is a homomorphism of groups, \( \mathrm{N} \vartriangleleft \mathrm{G},\mathrm{M} \vartriangleleft \mathrm{H} \), and \( \mathrm{f}\left( N\right) < \mathrm{M} \), then \( \mathrm{f} \) induces a homomorphism \( \bar{\mathrm{f}} : \mathrm{G}/\mathrm... | SKETCH OF PROOF. Consider the composition \( G\overset{f}{ \rightarrow }H\overset{\pi }{ \rightarrow }H/M \) and verify that \( N \subset {f}^{-1}\left( M\right) = \operatorname{Ker}{\pi f} \) . By Theorem 5.6 (applied to \( {\pi f} \) ) the map \( G/N \rightarrow H/M \) given by \( {aN} \mapsto \left( {\pi f}\right) \... | No |
Corollary 5.9. (Second Isomorphism Theorem) If \( \mathrm{K} \) and \( \mathrm{N} \) are subgroups of a group: \( \mathrm{G} \), with \( \mathrm{N} \) normal in \( \mathrm{G} \), then \( \mathrm{K}/\left( {\mathrm{N} \cap \mathrm{K}}\right) \cong \mathrm{{NK}}/\mathrm{N} \) . | PROOF. \( N \vartriangleleft {NK} = N \vee K \) by Theorem 5.3. The composition \( K\overset{ \subset }{ \rightarrow }{NK}\overset{\pi }{ \rightarrow } \) \( {NK}/N \) is a homomorphism \( f \) with kernel \( K \cap N \), whence \( \bar{f} : K/K \cap N \cong \operatorname{Im}f \) by Corollary 5.7. Every element in \( {... | Yes |
Corollary 5.10. (Third Isomorphism Theorem). If \( \mathrm{H} \) and \( \mathrm{K} \) are normal subgroups of a group \( \mathrm{G} \) such that \( \mathrm{K} < \mathrm{H} \), then \( \mathrm{H}/\mathrm{K} \) is a normal subgroup of \( \mathrm{G}/\mathrm{K} \) and \( \left( {\mathrm{G}/\mathrm{K}}\right) /\left( {\math... | PROOF. The identity map \( {1}_{G} : G \rightarrow G \) has \( {1}_{G}\left( K\right) < H \) and therefore induces an epimorphism \( I : G/K \rightarrow G/H \), with \( I\left( {aK}\right) = {aH} \) . Since \( H = I\left( {aK}\right) \) if and only if \( {a\varepsilon H} \), Ker \( I = \{ {aK} \mid {a\varepsilon H}\} =... | Yes |
Theorem 5.11. If \( \mathrm{f} : \mathrm{G} \rightarrow \mathrm{H} \) is an epimorphism of groups, then the assignment \( \mathrm{K} \mapsto \mathrm{f}\left( \mathrm{K}\right) \) defines a one-to-one correspondence between the set \( {\mathrm{S}}_{\mathrm{f}}\left( \mathrm{G}\right) \) of all subgroups \( \mathrm{K} \)... | SKETCH OF PROOF. By Exercise 2.9 the assignment \( K \mapsto f\left( K\right) \) defines a function \( \varphi : {S}_{f}\left( G\right) \rightarrow S\left( H\right) \) and \( {f}^{-1}\left( J\right) \) is a subgroup of \( G \) for every subgroup \( J \) of \( H \) . Since \( J < H \) implies \( \operatorname{Ker}f < {f... | No |
Corollary 5.12. If \( \mathrm{N} \) is a normal subgroup of a group \( \mathrm{G} \), then every subgroup of \( \mathrm{G}/\mathrm{N} \) is of the form \( \mathrm{K}/\mathrm{N} \), where \( \mathrm{K} \) is a subgroup of \( \mathrm{G} \) that contains \( \mathrm{N} \) . Furthermore, \( \mathrm{K}/\mathrm{N} \) is norma... | PROOF. Apply Theorem 5.11 to the canonical epimorphism \( \pi : G \rightarrow G/N \) . If \( N < K < G \), then \( \pi \left( K\right) = K/N \) . | Yes |
Corollary 6.4. The order of a permutation \( \sigma \in {\mathrm{S}}_{\mathrm{n}} \) is the least common multiple of the orders of its disjoint cycles. | PROOF. Let \( \sigma = {\sigma }_{1}\cdots {\sigma }_{r} \), with \( \left\{ {\sigma }_{i}\right\} \) disjoint cycles. Since disjoint cycles commute, \( {\sigma }^{m} = {\sigma }_{1}{}^{m}\cdots {\sigma }_{r}{}^{m} \) for all \( m \in \mathbf{Z} \) and \( {\sigma }^{m} = \left( 1\right) \) if and only if \( {\sigma }_{... | Yes |
Corollary 6.5. Every permutation in \( {\mathrm{S}}_{\mathrm{n}} \) can be written as a product of (not necessarily disjoint) transpositions. | PROOF. It suffices by Theorem 6.3 to show that every cycle is a product of transpositions. This is easy: \( \left( {x}_{1}\right) = \left( {{x}_{1}{x}_{2}}\right) \left( {{x}_{1}{x}_{2}}\right) \) and for \( r > 1,\left( {{x}_{1}{x}_{2}{x}_{3}\cdots {x}_{r}}\right) \) \( = \left( {{x}_{1}{x}_{r}}\right) \left( {{x}_{1}... | Yes |
Theorem 6.8. For each \( \mathrm{n} \geq 2 \), let \( {\mathrm{A}}_{\mathrm{n}} \) be the set of all even permutations of \( {\mathrm{S}}_{\mathrm{n}} \) . Then \( {\mathrm{A}}_{\mathrm{n}} \) is a normal subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) of index 2 and order \( \left| {\mathrm{S}}_{\mathrm{n}}\right| /2 = \m... | SKETCH OF PROOF OF 6.8. Let \( C \) be the multiplicative subgroup \( \{ 1, - 1\} \) of the integers. Define a map \( f : {S}_{n} \rightarrow C \) by \( \sigma \mapsto \operatorname{sgn}\sigma \) and verify that \( f \) is an epimorphism of groups. Since the kernel of \( f \) is clearly \( {A}_{n},{A}_{n} \) is normal ... | No |
Lemma 6.11. Let \( \mathrm{r},\mathrm{s} \) be distinct elements of \( \{ 1,2,\ldots ,\mathrm{n}\} \) . Then \( {\mathrm{A}}_{\mathrm{n}}\left( {\mathrm{n} \geq 3}\right) \) is generated by the 3-cycles \( \{ \left( \mathrm{{rsk}}\right) \mid 1 \leq \mathrm{k} \leq \mathrm{n},\mathrm{k} \neq \mathrm{r},\mathrm{s}\} \) ... | PROOF. Assume \( n > 3 \) (the case \( n = 3 \) is trivial). Every element of \( {A}_{n} \) is a product of terms of the form \( \left( {ab}\right) \left( {cd}\right) \) or \( \left( {ab}\right) \left( {ac}\right) \), where \( a, b, c, d \) are distinct elements of \( \{ 1,2,\ldots, n\} \) . Since \( \left( {ab}\right)... | Yes |
Lemma 6.12. If \( \mathrm{N} \) is a normal subgroup of \( {\mathrm{A}}_{\mathrm{n}}\left( {\mathrm{n} \geq 3}\right) \) and \( \mathrm{N} \) contains a 3-cycle, then \( \mathrm{N} = {\mathrm{A}}_{\mathrm{n}} \) . | PROOF. If \( \left( {rsc}\right) {\varepsilon N} \), then for any \( k \neq r, s, c,\left( {rsk}\right) = \left( {rs}\right) \left( {ck}\right) {\left( rsc\right) }^{2}\left( {ck}\right) \left( {rs}\right) \) \( = \left\lbrack {\left( {rs}\right) \left( {ck}\right) }\right\rbrack {\left( rsc\right) }^{2}{\left\lbrack \... | Yes |
Theorem 6.13. For each \( \mathrm{n} \geq 3 \) the dihedral group \( {\mathrm{D}}_{\mathrm{n}} \) is a group of order \( 2\mathrm{n} \) whose generators \( \mathrm{a} \) and \( \mathrm{b} \) satisfy:\n\n(i) \( {a}^{n} = \left( 1\right) ;{b}^{2} = \left( 1\right) ;{a}^{k} \neq \left( 1\right) \) if \( 0 < k < n \) ;\n\n... | SKETCH OF PROOF. Verify that \( a, b \in {D}_{n} \) as defined above satisfy (i) and (ii), whence \( {D}_{n} = \langle a, b\rangle = \left\{ {{a}^{i}{b}^{j} \mid 0 \leq i < n;j = 0,1}\right\} \) (see Theorem 2.8). Then verify that the \( {2n} \) elements \( {a}^{i}{b}^{j}\left( {0 \leq i < n;j = 0,1}\right) \) are all ... | Yes |
Theorem 7.3. If \( \left( {\mathrm{P},\left\{ {\pi }_{\mathrm{i}}\right\} }\right) \) and \( \left( {\mathrm{Q},\left\{ {\psi }_{\mathrm{i}}\right\} }\right) \) are both products of the family \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathbf{I}}\right\} \) of objects of a category \( \mathrm{C} \), the... | PROOF. Since \( P \) and \( Q \) are both products, there exist morphisms \( f : P \rightarrow Q \) and \( g : Q \rightarrow P \) such that the following diagrams are commutative for each \( {i\varepsilon I} \) :\n\n\n... | Yes |
Theorem 7.8. If \( \mathcal{C} \) is a concrete category, \( \mathbf{F} \) and \( {\mathbf{F}}^{\prime } \) are objects of \( \mathcal{C} \) such that \( \mathbf{F} \) is free on the set \( \mathrm{X} \) and \( {\mathrm{F}}^{\prime } \) is free on the set \( {\mathrm{X}}^{\prime } \) and \( \left| \mathrm{X}\right| = \... | PROOF OF 7.8. Since \( F,{F}^{\prime } \) are free and \( \left| X\right| = \left| {X}^{\prime }\right| \), there is a bijection \( f : X \rightarrow {X}^{\prime } \) and maps \( i : X \rightarrow F \) and \( j : {X}^{\prime } \rightarrow {F}^{\prime } \) . Consider the map \( {jf} : X \rightarrow {F}^{\prime } \) . Si... | Yes |
Theorem 7.10. Any two universal [resp. couniversal] objects in a category \( \mathcal{C} \) are equivalent. | PROOF. Let \( I \) and \( J \) be universal objects in \( \mathcal{C} \) . Since \( I \) is universal, there is a unique morphism \( f : I \rightarrow J \) . Similarly, since \( J \) is universal, there is a unique morphism \( g : J \rightarrow I \) . The composition \( g \circ f : I \rightarrow I \) is a morphism of \... | Yes |
Theorem 8.1. If \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) is a family of groups, then\n\n(i) the direct product \( \mathop{\prod }\limits_{{i \in I}}{\mathrm{G}}_{\mathrm{i}} \) is a group;\n\n(ii) for each \( \mathrm{k} \in \mathrm{I} \), the map \( {\pi }_{\mathrm{k}} : \... | PROOF. Exercise. | No |
Theorem 8.2. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of groups and \( \left\{ {{\varphi }_{\mathrm{i}} : \mathrm{H} \rightarrow {\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) a family of group homomorphisms. Then there is a unique homomorphi... | PROOF. By Introduction, Theorem 5.2, the map of sets \( \varphi : H \rightarrow \mathop{\prod }\limits_{{i \in I}}{G}_{i} \) given by \( \varphi \left( a\right) = {\left\{ {\varphi }_{i}\left( a\right) \right\} }_{i\epsilon I} \in \mathop{\prod }\limits_{{i\epsilon I}}{G}_{i} \) is the unique function such that \( {\pi... | Yes |
Theorem 8.4. If \( \\left\\{ {{G}_{i} \\mid {i\\varepsilon }\\mathbf{I}}\\right\\} \) is a family of groups, then\n\n(i) \( \\mathop{\\prod }\\limits_{{i \\in I}}{}^{\\mathrm{w}}{\\mathrm{G}}_{\\mathrm{i}} \) is a normal subgroup of \( \\mathop{\\prod }\\limits_{{i \\in I}}{\\mathrm{G}}_{\\mathrm{i}} \) ;\n\n(ii) for e... | PROOF. Exercise. | No |
Theorem 8.5. Let \( \\left\\{ {{\\mathrm{A}}_{\\mathrm{i}} \\mid \\mathrm{i}\\varepsilon \\mathrm{I}}\\right\\} \) be a family of abelian groups (written additively). If \( \\mathrm{B} \) is an abelian group and \( \\left\\{ {{\\psi }_{\\mathrm{i}} : {\\mathrm{A}}_{\\mathrm{i}} \\rightarrow \\mathrm{B} \\mid \\mathrm{i... | PROOF OF 8.5. Throughout this proof all groups will be written additively. If \( 0 \\neq \\left\\{ {a}_{i}\\right\\} \\varepsilon \\sum {A}_{i} \), then only finitely many of the \( {a}_{i} \) are nonzero, say \( {a}_{{i}_{1}},{a}_{{i}_{2}},\\ldots ,{a}_{{i}_{r}} \) . Define \( \\psi : \\sum {A}_{i} \\rightarrow B \) b... | Yes |
Theorem 8.9. Let \( \\left\\{ {{\\mathbf{N}}_{\\mathrm{i}} \\mid \\mathrm{i} \\in \\mathbf{I}}\\right\\} \) be a family of normal subgroups of a group \( \\mathbf{G} \) . \( \\mathbf{G} \) is the internal weak direct product of the family \( \\left\\{ {{\\mathbf{N}}_{\\mathrm{i}} \\mid \\mathrm{i} \\in \\mathbf{I}}\\ri... | ## PROOF. Exercise. | No |
Theorem 8.10. Let \( \left\{ {{\mathrm{f}}_{\mathrm{i}} : {\mathrm{G}}_{\mathrm{i}} \rightarrow {\mathrm{H}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) be a family of homomorphisms of groups and let \( \mathrm{f} = \prod {\mathrm{f}}_{\mathrm{i}} \) be the map \( \mathop{\prod }\limits_{{\mathrm{i}{... | PROOF. Exercise. | No |
Corollary 8.11. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) and \( \left\{ {{\mathrm{N}}_{\mathrm{i}} \mid \mathrm{i}\varepsilon \mathrm{I}}\right\} \) be families of groups such that \( {\mathrm{N}}_{\mathrm{i}} \) is a normal subgroup of \( {\mathbf{G}}_{\mathbf{i}} \) ... | PROOF. (i) For each \( i \), let \( {\pi }_{i} : {G}_{i} \rightarrow {G}_{i}/{N}_{i} \) be the canonical epimorphism. By Theorem 8.10, the map \( \prod {\pi }_{i} : \mathop{\prod }\limits_{{i \in I}}{G}_{i} \rightarrow \mathop{\prod }\limits_{{i \in I}}{G}_{i}/{N}_{i} \) is an epimorphism with kernel \( \mathop{\prod }... | Yes |
Theorem 9.1. If \( \mathrm{X} \) is a nonempty set and \( \mathrm{F} = \mathrm{F}\left( \mathrm{X}\right) \) is the set of all reduced words on \( \mathrm{X} \), then \( \mathrm{F} \) is a group under the binary operation defined above and \( \mathrm{F} = \langle \mathrm{X}\rangle \) . | SKETCH OF PROOF OF 9.1. Since 1 is an identity element and \( {x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}} \) has inverse \( {x}_{n}^{-{\delta }_{n}} \cdot \cdot \cdot {x}_{1}^{-{\delta }_{1}} \), we need only verify associativity. This may be done by induction and a tedious examination of cases or by the ... | Yes |
Theorem 9.2. Let \( \mathrm{F} \) be the free group on a set \( \mathrm{X} \) and \( \iota : \mathrm{X} \rightarrow \mathrm{F} \) the inclusion map. If \( \mathrm{G} \) is a group and \( \mathrm{f} : \mathrm{X} \rightarrow \mathrm{G} \) a map of sets, then there exists a unique homomorphism of groups \( \widetilde{\mat... | SKETCH OF PROOF OF 9.2. Define \( \bar{f}\left( 1\right) = e \) and if \( {x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}} \) is a nonempty reduced word on \( X \), define \( \bar{f}\left( {{x}_{1}{}^{{\delta }_{1}}\cdots {x}_{n}{}^{{\delta }_{n}}}\right) = f{\left( {x}_{1}\right) }^{{\delta }_{1}}f{\left( {x}... | No |
Every group \( \mathbf{G} \) is the homomorphic image of a free group. | Let \( X \) be a set of generators of \( G \) and let \( F \) be the free group on the set \( X \) . By Theorem 9.2 the inclusion map \( X \rightarrow G \) induces a homomorphism \( \bar{f} : F \rightarrow G \) such that \( x \mapsto x \in G \) . Since \( G = \langle X\rangle \), the proof of Theorem 9.2 shows that \( ... | Yes |
Theorem 9.5. (Van Dyck) Let \( \mathrm{X} \) be a set, \( \mathrm{Y} \) a set of (reduced) words on \( \mathrm{X} \) and \( \mathrm{G} \) the group defined by the generators \( \mathrm{x}\varepsilon \mathrm{X} \) and relations \( \mathrm{w} = \mathrm{e}\left( {\mathrm{w}\varepsilon \mathrm{Y}}\right) \) . If \( \mathrm... | PROOF OF 9.5. If \( F \) is the free group on \( X \) then the inclusion map \( X \rightarrow H \) induces an epimorphism \( \varphi : F \rightarrow H \) by Corollary 9.3. Since \( H \) satisfies the relations \( w = e\left( {w \in Y}\right), Y \subset \operatorname{Ker}\varphi \) . Consequently, the normal subgroup \(... | Yes |
Theorem 9.6. Let \( \left\{ {{\mathrm{G}}_{\mathrm{i}} \mid \mathrm{i} \in \mathrm{I}}\right\} \) be a family of groups and \( \mathop{\prod }\limits_{{i \in I}}{}^{ * }{\mathrm{G}}_{\mathrm{i}} \) their free product. If\n\n\( \left\{ {{\psi }_{\mathrm{i}} : {\mathrm{G}}_{\mathrm{i}} \rightarrow \mathrm{H} \mid \mathrm... | SKETCH OF PROOF. If \( {a}_{1}{a}_{2}\cdots {a}_{n} \) is a reduced word in \( \mathop{\prod }\limits_{{i \in I}}{}^{ * }{G}_{i} \) with \( {a}_{k} \in {G}_{{i}_{k}} \) , define \( \psi \left( {{a}_{1}\cdots {a}_{n}}\right) \) to be \( {\psi }_{{i}_{1}}\left( {a}_{1}\right) {\psi }_{{i}_{2}}\left( {a}_{2}\right) \cdots... | Yes |
Theorem 1.1. The following conditions on an abelian group \( \mathrm{F} \) are equivalent.\n\n(i) \( \mathrm{F} \) has a nonempty basis.\n\n(ii) \( \mathrm{F} \) is the (internal) direct sum of a family of infinite cyclic subgroups.\n\n(iii) \( \mathrm{F} \) is (isomorphic to) a direct sum of copies of the additive gro... | SKETCH OF PROOF OF 1.1. (i) \( \Rightarrow \) (ii) If \( X \) is a basis of \( F \), then for each \( {x\varepsilon X},{nx} = 0 \) if and only if \( n = 0 \) . Hence each subgroup \( \langle x\rangle \left( {x\varepsilon X}\right) \) is infinite cyclic (and normal since \( F \) is abelian). Since \( F = \langle X\rangl... | No |
Proposition 1.3. Let \( {\mathrm{F}}_{1} \) be the free abelian group on the set \( {\mathrm{X}}_{1} \) and \( {\mathrm{F}}_{2} \) the free abelian group on the set \( {\mathrm{X}}_{2} \) . Then \( {\mathrm{F}}_{1} \cong {\mathrm{F}}_{2} \) if and only if \( {\mathrm{F}}_{1} \) and \( {\mathrm{F}}_{2} \) have the same ... | SKETCH OF PROOF OF 1.3. If \( \alpha : {F}_{1} \cong {F}_{2} \), then \( \alpha \left( {X}_{1}\right) \) is a basis of \( {F}_{2} \) , whence \( \left| {X}_{1}\right| = \left| {\alpha \left( {X}_{1}\right) }\right| = \left| {X}_{2}\right| \) by Theorem 1.2. The converse is Theorem I.7.8. | Yes |
Theorem 1.4. Every abelian group \( \mathrm{G} \) is the homomorphic image of a free abelian group of rank \( \left| \mathrm{X}\right| \), where \( \mathrm{X} \) is a set of generators of \( \mathbf{G} \) . | PROOF. Let \( F \) be the free abelian group on the set \( X \) . Then \( F = \mathop{\sum }\limits_{{x \in X}}\mathbf{Z}x \) and rank \( F = \left| X\right| \) . By Theorem 1.1 the inclusion map \( X \rightarrow G \) induces a homomorphism \( \bar{f} : F \rightarrow G \) such that \( {1x} \mapsto {x\varepsilon G} \), ... | Yes |
Lemma 1.5. If \( \left\{ {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right\} \) is a basis of a free abelian group \( \mathrm{F} \) and \( \mathrm{a}\varepsilon \mathbf{Z} \), then for all \( \mathrm{i} \neq \mathrm{j}\left\{ {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{j} - 1},{\mathrm{x}}_{\mathrm{i}} + {... | PROOF. Since \( {x}_{j} = - a{x}_{i} + \left( {{x}_{j} + a{x}_{i}}\right) \), it follows that \( F = \left\langle {{x}_{1},\ldots ,{x}_{j - 1},{x}_{j} + }\right. \) \( \left. {a{x}_{i},{x}_{j + 1},\ldots ,{x}_{n}}\right\rangle \) . If \( {k}_{1}{x}_{1} + \cdots + {k}_{j}\left( {{x}_{j} + a{x}_{i}}\right) + \cdots + {k}... | Yes |
Corollary 1.7. If \( \mathrm{G} \) is a finitely generated abelian group generated by \( \mathrm{n} \) elements, then every subgroup \( \mathrm{H} \) of \( \mathrm{G} \) may be generated by \( \mathrm{m} \) elements with \( \mathrm{m} \leq \mathrm{n} \) . | PROOF OF 1.7. By Theorem 1.4 there is a free abelian group \( F \) of rank \( n \) and an epimorphism \( \pi : F \rightarrow G.{\pi }^{-1}\left( H\right) \) is a subgroup of \( F \), and therefore, free of rank \( m \leq n \) by Theorem 1.6. The image under \( \pi \) of any basis of \( {\pi }^{-1}\left( H\right) \) is ... | Yes |
Theorem 2.1. Every finitely generated abelian group \( \mathrm{G} \) is (isomorphic to) a finite direct sum of cyclic groups in which the finite cyclic summands (if any) are of orders \( {\mathrm{m}}_{1},\ldots ,{\mathrm{m}}_{\mathrm{t}} \), where \( {\mathrm{m}}_{1} > 1 \) and \( {\mathrm{m}}_{1}\left| {\mathrm{\;m}}_... | PROOF. If \( G \neq 0 \) and \( G \) is generated by \( n \) elements, then there is a free abelian group \( F \) of rank \( n \) and an epimorphism \( \pi : F \rightarrow G \) by Theorem 1.4. If \( \pi \) is an isomorphism, then \( G \cong F \cong \mathbf{Z} \oplus \cdots \oplus \mathbf{Z} \) ( \( n \) summands). If n... | Yes |
Theorem 2.2. Every finitely generated abelian group \( \mathrm{G} \) is (isomorphic to) a finite direct sum of cyclic groups, each of which is either infinite or of order a power of a prime. | SKETCH OF PROOF. The theorem is an immediate consequence of Theorem 2.1 and the following lemma. Another proof is sketched in Exercise 4. | No |
Lemma 2.3. If \( \mathrm{m} \) is a positive integer and \( \mathrm{m} = {\mathrm{p}}_{1}{}^{{\mathrm{n}}_{1}}{\mathrm{p}}_{2}{}^{{\mathrm{n}}_{2}}\cdots {\mathrm{p}}_{\mathrm{t}}{}^{{\mathrm{n}}_{\mathrm{t}}}\left( {{\mathrm{p}}_{1},\ldots ,{\mathrm{p}}_{\mathrm{t}}}\right. \) distinct primes and each \( {\mathrm{n}}_... | SKETCH OF PROOF. Use induction on the number \( t \) of primes in the prime decomposition of \( m \) and the fact that\n\n\[ \n{Z}_{rn} \cong {Z}_{r} \oplus {Z}_{n}\;\text{ whenever }\;\left( {r, n}\right) = 1, \]\n\nwhich we now prove. The element \( n = {n1\varepsilon }{Z}_{rn} \) has order \( r \) (Theorem I.3.4 (vi... | No |
Corollary 2.4. If \( \mathrm{G} \) is a finite abelian group of order \( \mathrm{n} \), then \( \mathrm{G} \) has a subgroup of order \( \mathrm{m} \) for every positive integer \( \mathrm{m} \) that divides \( \mathrm{n} \) . | SKETCH OF PROOF. Use Theorem 2.2 and observe that \( G \cong \mathop{\sum }\limits_{{i = 1}}^{k}{G}_{i} \) implies that \( \left| G\right| = \left| {G}_{1}\right| \left| {G}_{2}\right| \cdots \left| {G}_{k}\right| \) and for \( i \leq r,{p}^{r - i}{Z}_{{p}^{r}} \cong {Z}_{{p}^{i}} \) by Lemma 2.5 (v) below. | No |
Lemma 2.5. Let \( \mathrm{G} \) be an abelian group, \( \mathrm{m} \) an integer and \( \mathrm{p} \) a prime integer. Then each of the following is a subgroup of \( \mathrm{G} \) :\n\n(i) \( \mathrm{{mG}} = \{ \mathrm{{mu}} \mid \mathrm{u}\varepsilon \mathrm{G}\} \) ;\n\n(ii) \( \mathrm{G}\left\lbrack \mathrm{m}\right... | SKETCH OF PROOF. (i)-(iv) are exercises; the hypothesis that \( G \) is abelian is essential \( \left( {S}_{3}\right. \) provides counterexamples for (i)-(iii) and Exercise I.3.5 for (iv)). | No |
Corollary 2.7. Two finitely generated abelian groups \( \mathrm{G} \) and \( \mathrm{H} \) are isomorphic if and only if \( \mathrm{G}/{\mathrm{G}}_{\mathrm{t}} \) and \( \mathrm{H}/{\mathrm{H}}_{\mathrm{t}} \) have the same rank and \( \mathrm{G} \) and \( \mathrm{H} \) have the same invariant factors [resp. elementar... | PROOF. Exercise. | No |
Theorem 3.3. Ifa group \( \mathrm{G} \) satisfies either the ascending or descending chain condition on normal subgroups, then \( \mathrm{G} \) is the direct product of a finite number of indecomposable subgroups. | SKETCH OF PROOF. Suppose \( G \) is not a finite direct product of indecomposable subgroups. Let \( S \) be the set of all normal subgroups \( H \) of \( G \) such that \( H \) is a direct factor of \( G \) (that is, \( G = H \times {T}_{H} \) for some subgroup \( {T}_{H} \) of \( G \) ) and \( H \) is not a finite dir... | No |
Lemma 3.4. Let \( \mathrm{G} \) be a group that satisfies the ascending [resp. descending] chain condition on normal subgroups and \( \mathrm{f} \) a [normal] endomorphism of \( \mathrm{G} \) . Then \( \mathrm{f} \) is an automorphism if and only if \( \mathbf{f} \) is an epimorphism [resp. monomorphism]. | PROOF. Suppose \( G \) satisfies the ACC and \( f \) is an epimorphism. The ascending chain of normal subgroups \( \langle e\rangle < \operatorname{Ker}f < \operatorname{Ker}{f}^{2} < \cdots \) (where \( {f}^{k} = {ff}\cdots f \) ) must become constant, say \( \operatorname{Ker}{f}^{n} = \operatorname{Ker}{f}^{n + 1} \... | Yes |
Lemma 3.5. (Fitting) If \( \mathrm{G} \) is a group that satisfies both the ascending and descending chain conditions on normal subgroups and \( \mathrm{f} \) is a normal endomorphism of \( \mathrm{G} \), then for some \( \mathrm{n} \geq 1,\mathrm{G} = \operatorname{Ker}{\mathrm{f}}^{\mathrm{n}} \times \operatorname{Im... | PROOF. Since \( f \) is a normal endomorphism each \( \operatorname{Im}{f}^{k}\left( {k \geq 1}\right) \) is normal in \( G \) . Hence we have two chains of normal subgroups:\n\n\[ G > \operatorname{Im}f > \operatorname{Im}{f}^{2} > \cdots \;\text{ and }\;\langle e\rangle < \operatorname{Ker}f < \operatorname{Ker}{f}^{... | Yes |
Corollary 3.6. If \( \mathrm{G} \) is an indecomposable group that satisfies both the ascending and descending chain conditions on normal subgroup’s and \( \mathbf{f} \) is a normal endomorphism of \( \mathbf{G} \) , then either \( \mathrm{f} \) is nilpotent or \( \mathrm{f} \) is an automorphism. | PROOF. For some \( n \geq 1, G = \operatorname{Ker}{f}^{n} \times \operatorname{Im}{f}^{n} \) by Fitting’s Lemma. Since \( G \) is indecomposable either \( \operatorname{Ker}{f}^{n} = \langle e\rangle \) or \( \operatorname{Im}{f}^{n} = \langle e\rangle \) . The latter implies that \( f \) is nilpotent. If \( \operator... | Yes |
Let \( \mathrm{G}\left( { \neq \langle \mathrm{e}\rangle }\right) \) be an indecomposable group that satisfies both the ascending and descending chain conditions on normal subgroups. If \( {\mathrm{f}}_{1},\ldots ,{\mathrm{f}}_{\mathrm{n}} \) are normal nilpotent endomorphisms of \( \mathrm{G} \) such that every \( {\m... | SKETCH OF PROOF. Since each \( {f}_{{i}_{1}} + \cdots + {f}_{{i}_{r}} \) is an endomorphism that is normal (Exercise 8(c)), the proof will follow by induction once the case \( n = 2 \) is established. If \( {f}_{1} + {f}_{2} \) is not nilpotent, it is an automorphism by Corollary 3.6. Verify that the inverse \( g \) of... | No |
Theorem 3.8. (Krull-Schmidt) Let \( \mathrm{G} \) be a group that satisfies both the ascending and descending chain conditions on normal subgroups. If \( \mathrm{G} = {\mathrm{G}}_{1} \times {\mathrm{G}}_{2} \times \cdots \times {\mathrm{G}}_{\mathrm{s}} \) and \( \mathrm{G} = {\mathrm{H}}_{1} \times {\mathrm{H}}_{2} \... | SKETCH OF PROOF OF 3.8. Let \( P\left( 0\right) \) be the statement \( G = {H}_{1} \times \cdots \times {H}_{t} \) . For \( 1 \leq r \leq \min \left( {s, t}\right) \) let \( P\left( r\right) \) be the statement: there is a reindexing of \( {H}_{1},\ldots ,{H}_{t} \) such that \( {G}_{i} \cong {H}_{i} \) for \( i = 1,2,... | Yes |
Theorem 4.2. Let \( \mathrm{G} \) be a group that acts on a set \( \mathrm{S} \). (i) The relation on \( \mathrm{S} \) defined by \[ \mathrm{x} \sim {\mathrm{x}}^{\prime } \Leftrightarrow \mathrm{{gx}} = {\mathrm{x}}^{\prime }\;\text{ for some }\;\mathrm{g}\varepsilon \mathrm{G} \] is an equivalence relation. | PROOF. Exercise. | No |
Theorem 4.3. If a group \( \mathrm{G} \) acts on a set \( \mathrm{S} \), then the cardinal number of the orbit of \( \mathrm{x}\varepsilon \mathrm{S} \) is the index \( \left\lbrack {\mathrm{G} : {\mathrm{G}}_{\mathrm{x}}}\right\rbrack \) . | PROOF. Let \( g, h \in G \) . Since\n\n\[ \n{gx} = {hx} \Leftrightarrow {g}^{-1}{hx} = x \Leftrightarrow {g}^{-1}h \in {G}_{x} \Leftrightarrow h{G}_{x} = g{G}_{x},\n\]\n\nit follows that the map given by \( g{G}_{x} \mapsto {gx} \) is a well-defined bijection of the set of cosets of \( {G}_{x} \) in \( G \) onto the or... | Yes |
Corollary 4.4. Let \( \mathrm{G} \) be a finite group and \( \mathrm{K} \) a subgroup of \( \mathrm{G} \). (i) The number of elements in the conjugacy class of \( \mathrm{x}\varepsilon \mathrm{G} \) is \( \left\lbrack {\mathrm{G} : {\mathrm{C}}_{\mathrm{G}}\left( \mathrm{x}\right) }\right\rbrack \), which divides \( \l... | PROOF. (i) and (iii) follow immediately from the preceding Theorem and Lagrange’s Theorem I.4.6. Since conjugacy is an equivalence relation on \( G \) (Theorem 4.2), \( G \) is the disjoint union of the conjugacy classes \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{n} \), whence (ii) follows from (i). | Yes |
Theorem 4.5. If a group \( \mathrm{G} \) acts on a set \( \mathrm{S} \), then this action induces a homomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{S}\right) \), where \( \mathrm{A}\left( \mathrm{S}\right) \) is the group of all permutations of \( \mathrm{S} \) . | PROOF. If \( g \in G \), define \( {\tau }_{g} : S \rightarrow S \) by \( x \mapsto {gx} \) . Since \( x = g\left( {{g}^{-1}x}\right) \) for all \( x \in S \) , \( {\tau }_{g} \) is surjective. Similarly \( {gx} = {gy}\left( {x,{y\varepsilon S}}\right) \) implies \( x = {g}^{-1}\left( {gx}\right) = {g}^{-1}\left( {gy}\... | Yes |
Corollary 4.6. (Cayley) If \( \mathrm{G} \) is a group, then there is a monomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{G}\right) \) . Hence every group is isomorphic to a group of permutations. In particular every finite group is isomorphic to a subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) with \( \math... | PROOF. Let \( G \) act on itself by left translation and apply Theorem 4.5 to obtain a homomorphism \( \tau : G \rightarrow A\left( G\right) \) . If \( \tau \left( g\right) = {\tau }_{g} = {1}_{G} \), then \( {gx} = {\tau }_{g}\left( x\right) = x \) for all \( x \in G \) . In particular \( {ge} = e \), whence \( g = e ... | Yes |
Proposition 4.8. Let \( \mathrm{H} \) be a subgroup of a group \( \mathrm{G} \) and let \( \mathrm{G} \) act on the set \( \mathrm{S} \) of all left cosets of \( \mathrm{H} \) in \( \mathrm{G} \) by left translation. Then the kernel of the induced homomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{S}\right... | PROOF. The induced homomorphism \( G \rightarrow A\left( S\right) \) is given by \( g \mapsto {\tau }_{g} \), where \( {\tau }_{g} : S \rightarrow S \) and \( {\tau }_{g}\left( {xH}\right) = {gxH} \) . If \( g \) is in the kernel, then \( {\tau }_{g} = {1}_{S} \) and \( {gxH} = {xH} \) for all \( x \in G \) ; in partic... | Yes |
Corollary 4.9. If \( \mathrm{H} \) is a subgroup of index \( \mathrm{n} \) in a group \( \mathrm{G} \) and no nontrivial normal subgroup of \( \mathrm{G} \) is contained in \( \mathrm{H} \), then \( \mathrm{G} \) is isomorphic to a subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) . | PROOF. Apply Proposition 4.8 to \( H \) ; the kernel of \( G \rightarrow A\left( S\right) \) is a normal subgroup of \( G \) contained in \( H \) and must therefore be \( \langle e\rangle \) by hypothesis. Hence, \( G \rightarrow A\left( S\right) \) is a monomorphism. Therefore \( G \) is isomorphic to a subgroup of th... | Yes |
Corollary 4.10. If \( \mathrm{H} \) is a subgroup of a finite group \( \mathrm{G} \) of index \( \mathrm{p} \), where \( \mathrm{p} \) is the smallest prime dividing the order of \( \mathbf{G} \), then \( \mathbf{H} \) is normal in \( \mathbf{G} \). | PROOF. Let \( S \) be the set of all left cosets of \( H \) in \( G \) . Then \( A\left( S\right) \cong {S}_{p} \) since \( \left\lbrack {G : H}\right\rbrack = p \) . If \( K \) is the kernel of the homomorphism \( G \rightarrow A\left( S\right) \) of Proposition 4.8, then \( K \) is normal in \( G \) and contained in ... | Yes |
Lemma 5.1. If a group \( \mathrm{H} \) of order \( {\mathrm{p}}^{\mathrm{n}} \) (p prime) acts on a finite set \( \mathrm{S} \) and if \( {\mathrm{S}}_{0} = \left\{ {\mathrm{x}\varepsilon \mathrm{S} \mid \mathrm{{hx}} = \mathrm{x}}\right. \) for all \( \mathrm{h}\varepsilon \mathrm{H} \) ), then \( \left| \mathrm{S}\ri... | PROOF OF 5.1. An orbit \( \bar{x} \) contains exactly one element if and only if \( x \in {S}_{0} \) . Hence \( S \) can be written as a disjoint union \( S = {S}_{0} \cup {\bar{x}}_{1} \cup {\bar{x}}_{2} \cup \cdots \cup {\bar{x}}_{n} \), with \( \left| {\bar{x}}_{i}\right| > 1 \) for all \( i \) . Hence \( \left| S\r... | Yes |
Theorem 5.2. (Cauchy) If \( \mathrm{G} \) is a finite group whose order is divisible by a prime \( \mathrm{p} \) , then \( \mathbf{G} \) contains an element of order \( \mathbf{p} \) . | PROOF. (J. H. McKay) Let \( S \) be the set of \( p \) -tuples of group elements \( \left\{ {\left( {{a}_{1},{a}_{2},\ldots ,{a}_{p}}\right) \mid {a}_{i} \in G\text{and}{a}_{1}{a}_{2}\cdots {a}_{p} = e}\right\} \) . Since \( {a}_{p} \) is uniquely determined as \( {\left( {a}_{1}{a}_{2}\cdots {a}_{p - 1}\right) }^{-1} ... | Yes |
Corollary 5.3. A finite group \( \mathrm{G} \) is a p-group if and only if \( \left| \mathrm{G}\right| \) is a power of \( \mathrm{p} \) . | PROOF. If \( G \) is a \( p \) -group and \( q \) a prime which divides \( \left| G\right| \), then \( G \) contains an element of order \( q \) by Cauchy’s Theorem. Since every element of \( G \) has order a power of \( p, q = p \) . Hence \( \left| G\right| \) is a power of \( p \) . The converse is an immediate cons... | Yes |
Corollary 5.4. The center \( \mathrm{C}\left( \mathrm{G}\right) \) of a nontrivial finite \( \mathrm{p} \) -group \( \mathrm{G} \) contains more than one element. | PROOF. Consider the class equation of \( G \) (see page 91):\n\n\[ \left| G\right| = \left| {C\left( G\right) }\right| + \sum \left\lbrack {G : {C}_{G}\left( {x}_{i}\right) }\right\rbrack \]\n\nSince each \( \left\lbrack {G : {C}_{G}\left( {x}_{i}\right) }\right\rbrack > 1 \) and divides \( \left| G\right| = {p}^{n}\le... | Yes |
Lemma 5.5. If \( \mathrm{H} \) is a p-subgroup of a finite group \( \mathrm{G} \), then \( \left\lbrack {{\mathrm{N}}_{\mathrm{G}}\left( \mathrm{H}\right) : \mathrm{H}}\right\rbrack \equiv \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \) \( \left( {\;\operatorname{mod}\;\mathrm{p}}\right) \) . | PROOF. Let \( S \) be the set of left cosets of \( H \) in \( G \) and let \( H \) act on \( S \) by (left) translation. Then \( \left| S\right| = \left\lbrack {G : H}\right\rbrack \) . Also,\n\n\[ \n{xH\varepsilon }{S}_{0} \Leftrightarrow {hxH} = {xH}\text{ for all }{h\varepsilon H} \n\]\n\n\( \Leftrightarrow {x}^{-1}... | Yes |
Corollary 5.6. If \( \mathrm{H} \) is p-subgroup of a finite group \( \mathrm{G} \) such that \( \mathrm{p} \) divides \( \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \), then \( {\mathrm{N}}_{\mathrm{G}}\left( \mathrm{H}\right) \neq \mathrm{H} \) . | PROOF. \( 0 \equiv \left\lbrack {G : H}\right\rbrack \equiv \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \left( {\;\operatorname{mod}\;p}\right) \) . Since \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \geq 1 \) in any case, we must have \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack > ... | Yes |
Theorem 5.7. (First Sylow Theorem) Let \( \mathrm{G} \) be a group of order \( {\mathrm{p}}^{\mathrm{n}}\mathrm{m} \), with \( \mathrm{n} \geq 1,\mathrm{p} \) prime, and \( \left( {\mathrm{p},\mathrm{m}}\right) = 1 \) . Then \( \mathrm{G} \) contains a subgroup of order \( {\mathrm{p}}^{\mathrm{i}} \) for each \( 1 \le... | PROOF. Since \( p\left| \right| G \mid, G \) contains an element \( a \), and therefore, a subgroup \( \langle a\rangle \) of order \( p \) by Cauchy’s Theorem. Proceeding by induction assume \( H \) is a subgroup of \( G \) of order \( {p}^{i}\left( {1 \leq i < n}\right) \) . Then \( p \mid \left\lbrack {G : H}\right\... | Yes |
Corollary 5.8. Let \( \mathrm{G} \) be a group of order \( {\mathrm{p}}^{\mathrm{n}}\mathrm{m} \) with \( \mathrm{p} \) prime, \( \mathrm{n} \geq 1 \) and \( \left( {\mathrm{m},\mathrm{p}}\right) = 1 \) . Let \( \mathrm{H} \) be a p-subgroup of \( \mathrm{G} \) .\n\n(i) \( \mathrm{H} \) is a Sylow p-subgroup of \( \mat... | SKETCH OF PROOF. (i) Corollaries I.4.6 and 5.3 and Theorem 5.7. | No |
Theorem 5.9. (Second Sylow Theorem) If \( \mathrm{H} \) is a p-subgroup of a finite group \( \mathrm{G} \), and \( \mathrm{P} \) is any Sylow \( \mathrm{p} \) -subgroup of \( \mathrm{G} \), then there exists \( \mathrm{x}\varepsilon \mathrm{G} \) such that \( \mathrm{H} < \mathrm{{xP}}{\mathrm{x}}^{-1} \) . In particul... | PROOF. Let \( S \) be the set of left cosets of \( P \) in \( G \) and let \( H \) act on \( S \) by (left) translation. \( \left| {S}_{0}\right| \equiv \left| S\right| = \left\lbrack {G : P}\right\rbrack \left( {\;\operatorname{mod}\;p}\right) \) by Lemma 5.1. But \( p \nmid \left\lbrack {G : P}\right\rbrack \) ; ther... | Yes |
Theorem 5.10. (Third Sylow Theorem) If \( \mathrm{G} \) is a finite group and \( \mathrm{p} \) a prime, then the number of Sylow p-subgroups of \( \mathrm{G} \) divides \( \left| \mathrm{G}\right| \) and is of the form \( \mathrm{{kp}} + 1 \) for some \( \mathrm{k} \geq 0 \) . | PROOF. By the second Sylow Theorem the number of Sylow \( p \) -subgroups is the number of conjugates of any one of them, say \( P \) . But this number is \( \left\lbrack {G : {N}_{G}\left( P\right) }\right\rbrack \), a divisor of \( \left| G\right| \), by Corollary 4.4. Let \( S \) be the set of all Sylow \( p \) -sub... | Yes |
Theorem 5.11. If \( \mathrm{P} \) is a Sylow p-subgroup of a finite group \( \mathrm{G} \), then \( {\mathrm{N}}_{\mathrm{G}}\left( {{\mathrm{N}}_{\mathrm{G}}\left( \mathrm{P}\right) }\right) \) \( = {\mathrm{N}}_{\mathrm{G}}\left( \mathrm{P}\right) \) . | PROOF. Every conjugate of \( P \) is a Sylow \( p \) -subgroup of \( G \) and of any subgroup of \( G \) that contains it. Since \( P \) is normal in \( N = {N}_{G}\left( P\right), P \) is the only Sylow \( p \) -subgroup of \( N \) by Theorem 5.9. Therefore,\n\n\[ x \in {N}_{G}\left( N\right) \Rightarrow {xN}{x}^{-1} ... | Yes |
Corollary 6.2. If \( \mathrm{p} \) is an odd prime, then every group of order \( 2\mathrm{p} \) is isomorphic either to the cyclic group \( {\mathbf{Z}}_{2\mathrm{p}} \) or the dihedral group \( {\mathbf{D}}_{\mathrm{p}} \) . | PROOF. Apply Proposition 6.1 with \( q = 2 \) . If \( G \) is not cyclic, the conditions on \( s \) imply \( s \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \) . Hence \( G = \langle c, d\rangle ,\left| d\right| = 2,\left| c\right| = p \), and \( {dc} = {c}^{-1}d \) by Theorem I.3.4(v). Therefore, \( G \cong {D}_{p... | Yes |
Proposition 6.3. There are (up to isomorphism) exactly two distinct nonabelian groups of order 8: the quaternion group \( {\mathrm{Q}}_{8} \) and the dihedral group \( {\mathrm{D}}_{4} \). | SKETCH OF PROOF OF 6.3. Verify that \( {D}_{4} \cong {Q}_{8} \) (Exercise 10). If a group \( G \) of order 8 is nonabelian, then it cannot contain an element of order 8 or have every nonidentity element of order 2 (Exercise I.1.13). Hence \( G \) contains an element \( a \) of order 4. The group \( \langle a\rangle \) ... | No |
Proposition 6.4. There are (up to isomorphism) exactly three distinct nonabelian groups of order 12: the dihedral group \( {\mathrm{D}}_{6} \), the alternating group \( {\mathrm{A}}_{4} \), and a group \( \mathrm{T} \) generated by elements \( \mathrm{a},\mathrm{b} \) such that \( \left| \mathrm{a}\right| = 6,{\mathrm{... | SKETCH OF PROOF. Verify that there is a group \( T \) of order 12 as stated (Exercise 5) and that no two of \( {D}_{6},{A}_{4}, T \) are isomorphic (Exercise 6). If \( G \) is a non-abelian group of order 12, let \( P \) be a Sylow 3-subgroup of \( G \) . Then \( \left| P\right| = 3 \) and \( \left\lbrack {G : P}\right... | No |
Theorem 7.3. The direct product of a finite number of nilpotent groups is nilpotent. | PROOF. Suppose for convenience that \( G = H \times K \), the proof for more than two factors being similar. Assume inductively that \( {C}_{i}\left( G\right) = {C}_{i}\left( H\right) \times {C}_{i}\left( K\right) \) (the case \( i = 1 \) is obvious). Let \( {\pi }_{H} \) be the canonical epimorphism \( H \rightarrow H... | Yes |
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