A solutions manual for Algebra by Thomas W. Hungerford
3. Cyclic Groups
1. Let \(a\), \(b\) be elements of a group \(G\). Show that \(|a| = |a^{-1}|\); \(|ab| = |ba|\), and \(|a| = |cac^{-1}|\) for all \(c\in G\).
Proof. \(\quad\)By Theorem 2.8, the subgroup \(\langle a \rangle\) generated by \(a\) consists of all finite products \(a^n(n \in \mathbb{Z})\); thus \(\langle a \rangle = \langle a^{-1} \rangle\) and so \(|a| = |a^{-1}|\).
\(\quad\) Suppose that \(ab\) has a finite order \(m\). Since \[
\begin{gathered}
e = (ab)^m = abab\cdots ab = a(ba)^{m-1}b \Leftrightarrow \\
(ba)^{m-1} = a^{-1}e b^{-1} \Leftrightarrow (ba)^{m-1}b = a^{-1}
\Leftrightarrow (ba)^m = e,
\end{gathered}
\] \(|ba| \le |ab|\). Conversely, suppose that \(ba\) has a finite order. Similarly to the previous, we have that \(|ab| \le |ba|\) and so \(|ab| = |ba|\). Thus if one of \(|ab|\) and \(|ba|\) is finite, then the other is finite; otherwise both infinite, and by Theorem 3.2, every infinite cyclic group is isomorphic to the additive group \(\mathbb{Z}\), so \(|ab| = |ba| = \aleph_0\). Therefore, \(|ab| = |ba|\).
\(\quad\) By induction, \((cac^{-1})^m=cac^{-1}cac^{-1}\cdots cac^{-1} =ca^mc^{-1}\) for all \(n \in \mathbb{N}\). Suppose that \(cac^{-1}\) has finite order \(m\). Since \[
e = ca^mc^{-1} \Leftrightarrow c e c^{-1} = e = a^m,
\] \(|a| \le |cac^{-1}|\). Conversely, suppose that \(a\) has finite order \(m\). Since \[
e = a^m \Leftrightarrow c^{-1}c = a^m \Leftrightarrow e = ca^mc^{-1},
\] \(|a| \ge |cac^{-1}|\), and so \(|a| = |cac^{-1}|\). Thus if one of \(|a|\) and \(|cac^{-1}|\) is finite, then the other is finite; otherwise both infinite, and by Theorem 3.2, \(|a| = |cac^{-1}| = \aleph_0\). Therefore, \(|a| = |cac^{-1}|.\quad\square\)
2. Let \(G\) be an abelian group containing elements \(a\) and \(b\) of orders \(m\) and \(n\) respectively. Show that \(G\) contains an element whose order is the least common multiple of \(m\) and \(n\). [Hint: first try the case when \((m, n) = 1\).]
Proof. \(\quad\) Write prime factorizations of \(m\) and \(n\) as \[ m=\prod_{i}p_i^{\alpha_i} \text{ and } n=\prod_{i} p_i^{\beta_i}, \] and let \[ m'=\prod_{i: \alpha_i \geq \beta_i}p_i^{\alpha_i}, n'=\prod_{i: \beta_i> \alpha_i} p_i^{\beta_i}, a'=a^{m/m'} \text{ and } b'=b^{n/n'}. \] Note that \(m'\) divides \(m\), and \(n'\) divides \(n\), and \(m'\) and \(n'\) are relatively prime, and \(m'n'\) is the least common multiple of \(m\) and \(n\). We claim that the order of \(a'\) is \(m'\). Let \(k\) be the order of \(a'\). Since \(e=(a')^k=(a^{m/m'})^k=a^{mk/m'}\), \(m\) divides \(mk/m'\), and so \(m'\) divides \(k\). On the other hand, since \((a^{m/m'})^{m'}=a^m=e\), \(k\) divides \(m'\). So the order of \(a'\) is \(m'\). Similarly, the order of \(b'\) is \(n'\). Now, let the order of \(a'b' = r'\), we claim that \(r'\) is \(m'n'\). Since \[ (a'b')^{m'n'} = a^{(m/m')m'n'}b^{(n/n')m'n'} = a^{mn'}b^{nm'} = e, \] \(r'\) divides \(m'n'\), and since \[ e = (a'b')^{r'} = (a'b')^{r'm'} = a'^{r'm'}b'^{r'm'} = b'^{r'm'} = e, \] \(n'\) divides \(r'm'\). \(m'\) and \(n'\) are relatively prime, so \(n'\) divides \(r'\). Similarly, \(m'\) divides \(r'\); thus \(m'n'\) divides \(r\), and so \(r' = m'n'\). Therefore, \(G\) contains an element whose order is the least common multiple of \(m\) and \(n\).\(\quad\square\)
3. Let \(G\) be an abelian group of order \(pq\), with \((p,q) = 1\). Assume there exists \(a, b \in G\) such that \(|a| = p, |b| = q\) and show that \(G\) is cyclic.
Proof. \(\quad\)By Example 2, \(G\) contains an element \(a\) whose order is \(pq\). Thus it is clear that \(G=\langle a \rangle\).\(\quad\square\)
4. If \(f : G \to H\) is a homomorphism, \(a \in G\), and \(f(a)\) has finite order in \(H\), then \(|a|\) is infinite or \(|f(a)|\) divides \(|a|\).
Proof. \(\quad\)Let \(|f(a)|=n\in\mathbb{N}\). Suppose that \(|a|\) is finite \(m\). Since \(a^m=e_G\) and \(f(a)^n=e_H\), \(f(a^m)=f(a)^m=e_H\); thus \(n\mid m\).\(\quad\square\)
5. Let \(G\) be a multiplicative group of all nonsingular \(2\times 2\) matrices with rational entries. Show that \(a = \begin{pmatrix} 0 & -1\\ 0 & \phantom{-}1 \end{pmatrix}\) has order \(4\) and \(b = \begin{pmatrix} \phantom{-}0 & \phantom{-}1\\ -1 & -1 \end{pmatrix}\) has order \(3\), but \(ab\) has infinite order. Conversely, show that the additive group \(Z_2 \oplus \mathbb{Z}\) contains nonzero elements \(a, b\) of infinite order such that \(a + b\) has finite order.
Proof. \(\quad\)It is easy to show that \(a^4=I, b^3=I\) and \((ab)^n =\begin{pmatrix} 1 & n\\ 0 & 1 \end{pmatrix}\) for all \(n\in\mathbb{Z}\). Consider \((1,-1)\) and \((1,1)\in Z_2 \oplus \mathbb{Z}\). \(2n(1,-1)=(0, -2n)\), \((2n+1)(1,-1)=(1, -2n-1),\) \(2n(1,1)=(0, 2n)\), and \((2n+1)(1,1)=(1, 2n+1)\) for all \(n\in\mathbb{Z}\). Thus \((1,-1)\) and \((1,1)\) have infinite orders, but the order of \((1,-1)+(1,1)=(0,0)\) is 1.\(\quad\square\)
6. If \(G\) is a cyclic group of order \(n\) and \(k\mid n\), then \(G\) has exactly one subgroup of order \(k\).
Proof. \(\quad\)Since every subgroup of a cyclic group is cylic, consider cyclic subgroups of \(G\). Let \(G=\langle a\rangle\) and \(m=n/p\). Since \(a^{mk}=e_G\) and \(a^{mi}\neq e_G\) for \(i\) with \(0<i<k\), the order of \(\langle a^m\rangle\) is \(k\). On the other hand, if the order of \(\langle a^p\rangle\) is \(k\) where \(0<p<n\), then \(n\mid pk\). Since \(k\mid n\), \(n/k\mid p\), i.e., \(m\mid p\); thus \(a^p\in \langle a^m\rangle\), and so \(\langle a^m\rangle\) is exactly one subgroup of order \(k\).\(\quad\square\)
7. Let \(p\) be prime and \(H\) a subgroup of \(Z(p^\infty )\) (Exercise 1.10).
\(\quad\)(a) Every element of \(Z(p^\infty )\) has finite order \(p^n\) for some \(n \ge 0\).
\(\quad\)(b) If at least one element of \(H\) has order \(p^k\) and no element of \(H\) has order greater than \(p^k\), then \(H\) is the cyclic subgroup generated by \(\overline{1/p^k}\), whence \(H \cong Z_{p^k}\).
\(\quad\)(c) If there is no upper bound on the orders of elements in \(H\), then \(H = Z(p^\infty )\); [see Exercise-I.2].
\(\quad\)(d) The only proper subgroups of \(Z(p^\infty )\) are the finite cyclic groups \(C_n = \langle \overline{1/p^n}\rangle (n = 1,2,...)\). Furthermore, \(\langle 0\rangle = C_0 \le C_1 \le C_2 \le C_3 \le\cdots\).
\(\quad\)(e) Let \(x_1,x_2,...\) be elements of an abelian group \(G\) such that \(|x_1| = p, px_2 = x_1, px_3 = x_2,...,px_{n+1} = x_n,....\) The subgroup generated by the \(x_i(i \ge 1)\) is isomorphic to \(Z(p^\infty )\). [Hint: Verify that the map induced by \(x_i \mapsto \overline{1/p^i}\) is a well-defined isomorphism.]
Proof. \(\quad\)(a) Let \(x \in Z(p^\infty)\) and \(|x|=m\). \(x=\overline{a/p^i}\) for some \(a\) and \(i\in\mathbb{N}\); thus \(p^ix=a\sim 0\), and so \(m\mid p^i\). Therefore \(m=p^n\) for some \(n\le i\).
\(\quad\)(b) If \(a\in H\) and \(|a|=p^k\), then \(a=c/p^k\) and \((c,p)=1\) for some \(c\in\mathbb{N}\). Clearly \(c/p^k\) is a geneator of \(\langle\overline{1/p^k}\rangle\); thus \(a\in\langle\overline{1/p^k}\rangle\subset H\). If \(b\in H\) and \(|b|\neq p^k\), then \(b=d/p^n\) where \(n\le k\) and \((d,p)=1\) for some \(d\in\mathbb{N}\); thus \(b=dp^{k-n}/p^k\in\langle\overline{1/p^k}\rangle\), and so \(\langle\overline{1/p^k}\rangle=H\). \(H\) and \(Z_{p^k}\) are two cyclic groups of the same order. Therefore, \(H \cong Z_{p^k}\).
\(\quad\)(c) If \(a\in H\), then \(a=b/p^n\) and \((b,p)=1\) for some \(b,n\in\mathbb{N}\); thus \(|a|=p^n\), and so \(\overline{1/p^m}\in H\) for all \(m\in\mathbb{N}\) with \(0<m\le n\). If there is no upper bound on the orders of elements in \(H\), then \(\{\overline{1/p^m}\mid m\in\mathbb{N^\ast}\}\subset H\), thus \(H=Z(p^\infty)\).
\(\quad\)(d) Let \(H<Z(p^\infty)\). By (c), if \(H\) is infinite then \(H=Z(p^\infty)\) and if finite \(S\subset Z(p^\infty)\) then \(\langle S\rangle = C_n\) for some \(n\in\mathbb{N^\ast}\).
\(\quad\)(e) Let \(H\) be the subgroup generated by the \(x_i(i \ge 1)\) and \(f\) be the function of \(H\) to \(Z(p^\infty)\) given by \(x_i \mapsto \overline{1/p^i}\). Notice that \(|x_n|=p^n\), \(ax_i=x_1\) for all \(i\) and some \(a\in\mathbb{N}^\ast\), and every element in H is a finite sum of multiples of \(x_i\). Thus if \(a\in H\), then \(a=bx_k\) for some \(b, k\in\mathbb{N}^\ast\) where \((b,p)=1\), and so \(a\in\langle x_k\rangle\). Moreover, if finite \(S\subset H\) then \(\langle S\rangle=\langle x_k\rangle\) for some \(k\in\mathbb{N}^\ast\) and if infinite \(T\subset H\) then \(\langle T\rangle=H\). It is easy to see that \(f(ab)=f(a)f(b)\) and \(f\) is bijective.\(\quad\square\)
8. A group that has only a finite number of subgroups must be finite.
Proof. \(\quad\)Let \(G\) be a infinite group. If there is an infinite subgroup \(\langle a \rangle\) of \(G\), then \(\langle a \rangle\cong\mathbb{Z}\), and so, for all \(m\in\mathbb{Z}\), there is a subgroup \(\langle ma \rangle\), i.e., there are infinitely many subgroups. If \(\langle a \rangle\) is finite for all \(a\in G\), then there are infinitely many finite subgroups of \(G\); otherwise \(\bigcup_{a\in G}\langle a\rangle=G\) is finite, a contradiction. Thus a infinite group has infinitely many subgroups. Therefore, a group that has only a finite number of subgroups must be finite.\(\quad\square\)
9. If \(G\) is an abelian group, then the set \(T\) of all elements of \(G\) with finite order is a subgroup of \(G\). [Compare Exercise 5.]
Proof. \(\quad\)\(e_G\in T\). If \(a\) and \(b\in T\) then \(e_G=ma=nb\) for some \(m,n\in \mathbb{Z}\); thus \(mna+mnb=e_G=mn(a+b)\), and so \(ab\in T\). If \(a\in T\), then clearly \(-a\in T\).\(\quad\square\)
10. An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups.
Proof. \(\quad\)Let \(G\) be a infinite group and \(a\in G\). If \(G=\langle a\rangle\), then for all \(m\in\mathbb{Z}\) with \(m\neq 0\), \(G\cong\mathbb{Z}\cong\langle ma\rangle\) and each of proper subgroups of \(G\) is cyclic. If \(G\) is isomorphic to each of its proper subgroup, then \(\langle b\rangle\cong G\) for each proper cyclic subgroup \(\langle b\rangle\) of \(G\). Thus \(G\) is cyclic. Notice that if there is a proper subgroup \(H<G\) then there is a proper cyclic subgroup of \(G\); otherwise, for all \(c\in H\), \(\langle c\rangle\neq H\) and so \(\langle c\rangle=\langle e_G \rangle\), a contradiction.\(\quad\square\)