A solutions manual for Algebra by Thomas W. Hungerford
2. Homomorphisms and Subgroups
1. If \(f : G \to H\) is a homomorphism of groups, then \(f(e_G) = e_H\) and \(f(a^{-1}) = f(a)^{-1}\) for all \(a \in G\). Show by example that the first conclusion may be false if \(G, H\) are monoids that are note groups.
Proof. \(\quad\)\(f(ae_G)=f(a)=f(a)f(e_G)\) \(=\) \(f(e_Ga)=f(e_G)f(a)\). Thus \(f(e_G)=e_H.\) \(f(aa^{-1})=f(e_G)=e_H\) \(=\) \(f(a)f(a^{-1})=f(a^{-1}a)\) \(=f(a^{-1})f(a)\). Thus \(f(a^{-1})=f(a)^{-1}\). Consider \(Z_3\) and \(Z_6\) under mutiplication. They form abelian monoids. Let \(f:Z_3\to Z_6\) be the function given by \(\{(0,0), (1,4), (2,2)\}\). \(f(ab)=f(a)f(b)\) for each \(a,b\in Z_3\), but \(f({1})\neq {1}\).\(\quad\square\)
2. A group \(G\) is abelian if and only if the map \(G\to G\) given by \(x \to x^{-1}\) is an automorphism.
Proof. \(\quad\)Let \(a,b\in G\) and let \(f\) be the map. Remind that by Exercise 1-11 (iii), \(G\) is abelian if and only if \((ab)^{-1}=a^{-1}b^{-1}\). Case \(\Rightarrow\). If \(G\) is abelian, then \(f(ab)=(ab)^{-1}=a^{-1}b^{-1}=f(a)f(b)\). Thus \(f\) is a homomorphism. There is \(a^{-1}\) such that \(f(a^{-1})\) = \(a\) for every \(a\). Thus \(f\) is surjective. If \(f(a)\neq f(b)\), then: \[ a^{-1}\neq b^{-1} \Leftrightarrow e\neq b^{-1}a \Leftrightarrow a\neq b. \] Thus \(f\) is injective. Therefore, \(f\) is an automorphism. Case \(\Leftarrow\). If \(f\) is an automorphism, then \(f(ab)=(ab)^{-1}=f(a)f(b)=a^{-1}b^{-1}\). \(G\) is abelian.\(\quad\square\)
3. Let \(Q_8\) be the group (under ordinary matrix multiplication) generated by the complex matrices \(A=\begin{pmatrix} \phantom{-}0 & 1\\ -1 & 0 \end{pmatrix}\) and \(B=\begin{pmatrix} 0 & i\\ i & 0 \end{pmatrix}\), where \(i^2 = -1\). Show that \(Q_8\) is a non-abelian group of order 8. \(Q_8\) is called the quaternion group. [Hint: Observe that \(BA = A^3B\), whence every element of \(Q_8\) is of the form \(A^iB^j\). Note also that \(A^4 = B^4 = I\), where \(I = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\) is the identity element of \(Q_8\).]
Proof. \(\quad\)\(A^2=\begin{pmatrix} \phantom{-}0 & -1\\ -1 & \phantom{-}0 \end{pmatrix}=-I\), \(A^3=-A, A^4=I\). \(B^2=-I, B^3=-B, B^4=I\). \(AB=\begin{pmatrix} i & \phantom{-}0\\ 0 & -i \end{pmatrix}\) \(\neq\) \(\begin{pmatrix} -i & 0\\ \phantom{-}0 & i \end{pmatrix}\) \(=BA\). \(BA=-AB=-AIB=A^3B\). By \(BA=A^3B\), we have \(B^mA^n=A^pB^m\) for some \(p\). Thus every element of \(Q_8\) is of the form \(A^iB^j\). From this, we have \(Q_8=\{I,A,A^2,A^3,B,AB,A^2B,A^3B\}\).\(\quad\square\)
4. Let \(H\) be the group (under matrix multiplication) of real matrices generated by \(C = \begin{pmatrix} \phantom{-}0 & 1\\ -1 & 0 \end{pmatrix}\) and \(D = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\). Show \(H\) is a non-abelian group of order 8 which is not isomorphic to the quaternion group of Exercise 3, but is isomorphic to the group \(D_4^{\ast }\).
Proof. \(\quad\)Similarly to Exercise 3, we have \(C^2=-I,C^3 =-C\), \(C^4=D^2 =I\), \(DC\neq CD, DC=C^3D\), and \(H=\{I,C,C^2, C^3, D,CD,-D,-CD\}\). Define the map \(f:D^\ast_4\to H\) given by \[ \begin{pmatrix} I&C&C^2& C^3& D&CD&-D&-CD\\ I&R&R^2&R^3&T_x&T_{1,3}&T_y&T_{2,4} \end{pmatrix}. \] \(f\) is an isomorphism between \(D_4^\ast\) and \(H\). Let \(S=\{A, -A, B, A^2B,AB, A^3B\}\subset Q_8\). For all \(X\in S\), \(X^4=I\) and \(X^2\neq I\). But \(H\) has only \(2\) elements with the property, \(C\) and \(C^3\). Thus \(H\) is not isomorphic to \(Q_8\). \(\quad\square\)
5. Let \(S\) be a nonempty subset of a group \(G\) and define a relation on \(G\) by \(a \sim b\) if and only if \(ab^{-1} \in S\). Show that \(\sim\) is an equivalence relation if and only if \(S\) is a subgroup of \(G\).
Proof. \(\quad\)Case \(\Rightarrow\). Suppose that \(\sim\) is an equivalence relation. For all \(a\in G\), \(a\sim a=aa^{-1}=e\in S\). If \(a\in S\), then \(a=ae^{-1}\in S\), thus \(a\sim e\) and \(e\sim a\). If \(a\) and \(b\in S\), then \(e\sim a\) and \(e\sim b\), thus \(a\sim b\), and so \(ab^{-1}\in S\). By Theorem 2.5, \(S\) is a subgroup.
\(\quad\)Case \(\Leftarrow\). Suppose that \(S\) is a subgroup of \(G\). For all \(a\in G\), since \(aa^{-1}=e\in S\), \(a\sim a\). If \(a\sim b\), then \(ab^{-1}\in S\), and since \(S\) is a group, \((ab^{-1})^{-1}=ba^{-1}\in S\), thus \(b\sim a\). If \(a\sim b\) and \(b\sim c\), then \(ab^{-1}\in S\) and \(bc^{-1}\in S\), thus \(ab^{-1}bc^{-1}=ac^{-1}\in S\), and so \(a\sim c\).\(\quad\square\)
6. A nonempty finite subset of a group is a subgroup if and only if it is closed under the product in \(G\).
Proof. \(\quad\)Let \(S\) a finite subset of a group \(G\). Case \(\Rightarrow\). Clear. Case \(\Leftarrow\). If \(a\in S\), then \(a^n\) for \(n>0\) is a product of the elements of \(S\). Since \(S\) is finite, \(a^m=e\) for some \(m\), thus \(e\in S\), and so \(a^{-1}\in S\).\(\quad\square\)
7. If \(n\) is a fixed integer, then \(\{ kn \mid k \in \mathbb{Z}\} \subset \mathbb{Z}\) is an additive subgroup of \(\mathbb{Z}\), which is isomorphic to \(\mathbb{Z}\).
Proof. \(\quad\)Let \(K\) be the set. Clearly, \(K\) is an additive subgroup of \(\mathbb{Z}\), and the function \(f:\mathbb{Z}\to K\) given by \(n\mapsto kn\) is an isomorphism.\(\quad\square\)
8. The set \(\{\sigma \in S_n \mid \sigma (n) = n\}\) is a subgroup of \(S_n\), which is isomorphic to \(S_{n-1}\).
Proof. \(\quad\)There is an isomorphism \(f:S_{n-1}\to S_n\) given by \(\sigma\mapsto \sigma\cup\{(n,n)\}\).\(\quad\square\)
9. Let \(f:G\to H\) be a homomorphism of groups, A a subgroup of \(G\), and \(B\) a subgroup of \(H\).
\(\quad\)(a) \(\text{Ker }f\) and \(f^{-1}(B)\) are subgroups of \(G\).
\(\quad\)(b) \(f(A)\) is a subgroup of \(H\).
Proof. \(\quad\)(a) \(e_G\in\text{Ker }f\). If \(a,b\in\text{Ker }f\), then \(f(ab)=f(a)f(b)=e_H\), thus \(ab\in\text{Ker }f\). If \(a\in\text{Ker }f\), then \(f(a^{-1})=f(a)^{-1}=e_H\), thus \(a^{-1}\in \text{Ker }f\). Therefore, \(\text{Ker }f\) is a subgroup of \(G\). \(e_G\in f^{-1}(B)\). If \(a,b\in f^{-1}(B)\), then \(f(ab)=f(a)f(b)\in B\), thus \(ab\in f^{-1}(B)\). If \(a\in f^{-1}(B)\), then: \[
f(a)^{-1}=f(a^{-1})\in B \Rightarrow a^{-1}\in f^{-1}(B).
\] Therefore, \(f^{-1}(B)\) is a subgroup of \(G\).
\(\quad\)(b) \(g_H\in f(A)\). There is \(a\in A\) such that \(f(a)=b\) for every \(b\in f(A)\), thus \(f(a^{-1})=f(a)^{-1}=b^{-1}\). Similarly, there is \(a\) and \(b\in A\) such that \(f(a)=c\) and \(f(b)=d\) for every \(c\) and \(d\in f(A)\), thus \(cd=f(a)f(b)\in f(A)\). \(\quad\square\)
10. List all subgroups of \({Z}_2 \bigoplus {Z}_2\). Is \({Z}_2 \bigoplus Z_2\) isomorphic to \(Z_4\)?
Proof. \(\quad\)\(\langle(0, 0)\rangle\), \(\langle(1, 1)\rangle\), \(\langle(1, 0)\rangle\), \(\langle(0, 1)\rangle, {Z}_2 \bigoplus {Z}_2\). Similarly to Exercise 4, \((0, 1)\) and \((1, 0)\in {Z}_2 \bigoplus {Z}_2\) have order \(2\), but only \(\overline{2}\in Z_4\) has order \(2\). Therefore, \({Z}_2 \bigoplus {Z}_2\) is not isomorphic to \(Z_4\).\(\quad\square\)
11. If \(G\) is a group, then \(C =\{a\in G\mid ax=xa \text{ for all }x\in G\}\) is an abelian subgroup of \(G\). \(C\) is called the center of \(G\).
Proof. \(\quad\)Let \(x\in G\) and \(a,b\in C\). \(ax=xa \Rightarrow axa^{-1}=x\) \(\Rightarrow a^{-1}axa^{-1}=a^{-1}x \Rightarrow xa^{-1}=a^{-1}x\) \(\Rightarrow a^{-1}\in C\). \(e_G\in C\). \(ab=ba\). \(xab=axb=abx\Rightarrow ab\in C\).\(\quad\square\)
12. The group \(D_4^{\ast }\) is not cyclic, but can be generated by two elements. The same is true of \(S_n\) (nontrivial). What is the minimal number of generators of the additive group \(\mathbb{Z} \bigoplus \mathbb{Z}\)?
Proof. \(\quad\)\(\mathbb{Z} \bigoplus \mathbb{Z}=\langle(1, 0), (0, 1)\rangle\).\(\quad\square\)
13. If \(G = \langle a \rangle\) is a cyclic group and \(H\) is any group, then every homomorphism \(f : G \to H\) is completely determined by the element \(f(a) \in H\).
Proof. \(\quad\)If \(b\in G\), then \(b=a^n\) for some \(n\in\mathbb{Z}\). If \(c\in H\), then \(c=f(a^m)=f(a)f(a^{m-1})=f(a)^m\) for some \(m\in\mathbb{Z}\). Thus every homomorphism of a cyclic group \(G\) is given by \(a^n\mapsto f(a)^n\) for \(n\in\mathbb{Z}\).\(\quad\square\)
14. The following cyclic subgroups are all isomorphic: the multiplicative group \(\langle i \rangle\) in \(C\), the additive group \({Z}_4\) and the subgroup \(\left< \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix} \right>\) of \(S_4\).
Proof. \(\quad\)They are all cyclic groups of order 4. If we map between \(i, 1\) and \(\left< \begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix} \right>\), then isomorphisms are determined.\(\quad\square\)
15. Let \(G\) be a group and \(\text{Aut }G\) the set of all automorphisms of \(G\).
\(\quad\)(a) \(\text{Aut }G\) is a group with composition of functions as a binary operation. [Hint: \(1_G \in \text{Aut }G\) is an identity; inverses exist by Theorem 2.3.]
\(\quad\)(b) \(\text{Aut } \mathbb{Z} \cong {Z}_2\) and \(\text{Aut } Z_6 \cong Z_2\); \(\text{Aut }Z_8 \cong Z_2 \bigoplus Z_2\); \(\text{Aut } Z_p \cong Z_{p-1}\) (\(p\) prime).
\(\quad\)(c) What is the \(\text{Aut }Z_n\) for arbitrary \(n \in \mathbb{N}^*\)?
Proof. \(\quad\)(a) \(1_G\in\text{Aut }G\). A composition of bijective homomorphisms are a bijective homomorphism. By theorem 2.3, an isomorphism has an inverse.
\(\quad\)(b) Notice that every abelian group has an automorphism given by \(x\mapsto x^{-1}\). \(\text{Aut } \mathbb{Z}=\{1_{\mathbb{Z}}, -1_{\mathbb{Z}}\}\). \(\{(1_{\mathbb{Z}},0),(-1_{\mathbb{Z}},1)\}\) is an isomorphism. Similarly to the previous, \(\text{Aut } {Z_6}=\{1_{Z_6}, -1_{Z_6}\}\). \(\{(1_{Z_6}, 0), (-1_{Z_6}, 1)\}\) is an isomorphism. Let \(x\in S=\{1,3,5,7\}\subset Z_8\). Since \(x\) and \(8\) are relatively prime, \(xi\equiv xj\) (mod \(8\)) if and only if \(i\equiv j\) (mod \(8\)), thus every \(xn\) (mod \(8\)) for \(0\le n<8\) is different from each other, and so \(x\) is a generator of \(Z_8\). Hence we have \(\text{Aut }Z_8=\) \(\{1_{Z_8},3_{Z_8},5_{Z_8},7_{Z_8}\}\) where each element given by \(n\mapsto n\), \(n\mapsto 3n\), \(n\mapsto 5n\), and \(n\mapsto 7n\) respectively. It is easily seen that \(\{((0,0), 1_{Z_8}), ((0,1), 3_{Z_8}),\) \(((1,0),5_{Z_8}), ((1,1),7_{Z_8})\}\) is an isomorphism. Similarly to the previous, \(\text{Aut } Z_p=\{1_{Z_p}, 2_{Z_p},\ldots,(p-1)_{Z_p}\}\). If \(p=2\), then \(\text{Aut } Z_p\) is a cyclic group of order 1. If \(p>2\), then \(p\) has a primitive root modulo \(n\). Thus if we let \(r\) be a primitive root modulo \(n\) of \(p\), then \(r_{Z_p}\) is a generator of \(\text{Aut } Z_p\). Therefore, a function of \(\text{Aut } Z_p\) into \(Z_{p-1}\) given by \(\overline{r^n}_{Z_p}\mapsto \overline{n}\) is an isomorphism.
\(\quad\)(c) \((a, n)=1\) if and only if \(\langle a\rangle=Z_n\) for \(0< a<n\), and the function of \(Z_n\) given by \(\overline{x}\mapsto \overline{ax}\) for \(x\in\mathbb{Z}\) is an automorphism. Consider \(S=\{a\in Z_n\mid (a,n) = 1\}\) under multiplication. \(0\notin S\). \(1\in S\). Let \(a,b\in S\). \((ab, n) = 1\), thus \(ab\in S\). Since \(S\) is finite, \(a^p\equiv a^q\) (mod \(n\)) for some \(0<p<q\), and so \(a^{q-p-1}a\) \(\equiv\) \(aa^{p-q-1}\equiv 1\) (mod \(n\)). Since \(a\neq 0\), \({a^{q-p-1}}\in S\). Therefore, \(S\) is a group, and so is \(\text{Aut }Z_n\).\(\quad\square\)
16. For each prime \(p\) the additive subgroup \(\mathbb{Z}(p^\infty)\) of \(\mathbb{Q}/\mathbb{Z}\) (Exercise 1.10) is generated by the set \(\{\overline{1/p^n} \mid n\in \mathbb{N}^\ast\}\).
Proof. \(\quad\)Let a set \(S\) which is generated by \(\{\overline{1/p^n}\mid n\in \mathbb{N}^\ast\}\). Since is clear that \(S\subset \mathbb{Z}(p^\infty)\), it suffices to show that \(S\supset \mathbb{Z}(p^\infty)\). Let \(a\in\mathbb{Z}(p^\infty)\). \(a=\overline{m/p^n}=m(\overline{1/p^n})\) for some \(m,n\in\mathbb{Z}\), thus \(a\in S\).\(\quad\square\)
17. Let \(G\) be an abelian group and let \(H,K\) be subgroups of \(G\). Show that the join \(H \vee K\) is the set \(\{ab \mid a \in H, b \in K\}\). Extend this result to any finite number of subgroups of \(G\).
Proof. \(\quad\)Let \(S\) be the set. Since it is clear that \(H\cup K\subset S\) \(\subset H \vee K\), it suffices to show that \(S\) is a group. \(e_G\in S\). If \(ab\) and \(cd\in S\) for some \(a,b\in H\) and \(c,d\in K\), then \(ab\in H\) and \(cd\in K\), thus \(abcd\in S\). If \(ab\in S\) for some \(a\in H\) and \(b\in K\), then \(b^{-1}a^{-1}=a^{-1}b^{-1}\in S\). Therefore, \(S\) is a group.
\(\quad\)Let \(G\) be an abelian group, and let \(H=\{H_i\in G\mid i\in I\}\) be a nonempty finite family of subgroups of \(G\). It is easily seen by induction that the join \(H_1\vee H_2\vee\cdots\vee H_n\) is the set \(\{a_1a_2\cdots a_n\mid a_i\in H_i\}\).\(\quad\square\)
18. \(\quad\)(a) Let \(G\) be a group and \(\{H_i \mid i \in I\}\) a family of subgroups. State and prove a condition that will imply that \(\bigcup_{i \in I}H_i\) is a subgroup, that is, that \(\bigcup_{i \in I}H_i = \langle\bigcup_{i \in I}H_i\rangle\).
\(\quad\)(b) Give an example of a group \(G\) and a family of subgroups \(\{H_i \mid i \in I\}\) such that \(\bigcup_{i \in I}H_i \ne \langle\bigcup_{i \in I}H_i\rangle\).
Proof. \(\quad\)(a) \(\bigcup_{i \in I}H_i = \langle\bigcup_{i \in I}H_i\rangle\) if and only if \(\bigcup_{i \in I}H_i\) is closed under the operation of \(G\). \(\bigcup_{i \in I}H_i\) has an identity element, and an inverse of each element, thus if \(\bigcup_{i \in I}H_i\) is closed under the operation of \(G\), \(\bigcup_{i \in I}H_i\) is the smallest group which includes itself.
\(\quad\)(b) \(\bigcup \{Z_2, Z_3\}\neq \langle \bigcup \{Z_2, Z_3\}\rangle=\mathbb{Z}\).\(\quad\square\)
19. (a) The set of all subgroups of a group \(G\), partially ordered by set theoretic inclusion, forms a complete lattice (Introduction, Exercise 7.1 and 7.2) in which the g.l.b. of \(\{H_i \mid i \in I\}\) is \(\bigcap_{i \in I}H_i\) and the l.u.b. is \(\langle\bigcup_{i \in I}H_i\rangle\).
\(\quad\)(b) Exhibit the lattice of subgroups of the groups \(S_3, D_4^{\ast }, Z_6, Z_{27}\), and \(Z_{36}\).
Proof. \(\quad\)(a) Let \(S\) be the set of all subgroups of a group \(G\). \(\bigcap_{i \in I}H_i\) is the g.l.b of \(\{H_i \mid i \in I\}\) in \(P(G)\) under set inclusion, and also \(\bigcap_{i \in I}H_i\in S\). Consider a set \(T=\{t\in S\mid \bigcup_{i \in I}H_i\subset t\}\). Since \(G\in T\), \(T\) is nonempty, and so \(T\) has a nonempty minimal element under set inclusion. Let \(A\) and \(B\) be minimal elements of \(T\). If \(A\ne B\), then \(A\cap B\in T\) and \(A\cap B\subsetneq A\), a contradiction. Thus \(T\) has the smallest element \(A\), and by definition, \(A=\langle\bigcup_{i \in I}H_i\rangle\).
\(\quad\)(b) \(\left< \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \right>\) \(<\left< \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} \right>\), \(\left< \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} \right>\), \(\left< \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} \right> < S_3\). \(\left< \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \right>\) \(<\left< \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} \right> < S_3\).
\(\quad\)\(\langle 0\rangle<\langle 3\rangle<Z_6\). \(\langle 0\rangle<\langle 2\rangle<Z_6\) Similarly to \(D_4^{\ast }, Z_{27}\), and \(Z_{36}\).\(\quad\square\)