A solutions manual for Algebra by Thomas W. Hungerford
4. Cosets and Counting
1. Let \(G\) be a group and \(\{H_i\mid i\in I\}\) a family of subgroups. Then for any \(a\in G\), \((\bigcap_i H_i)a=\bigcap_i H_ia\).
Proof. \(\quad\)Let \(m\in(\bigcap_i H_i)a\). \(m=ha\) for some \(h\in \bigcap_i H_i\), and so \(ha\in\bigcap_i H_ia\). Conversely, let \(n\in\bigcap_i H_ia\). There is \(h_i\in H_i\) such that \(h_ia=m\) for all \(i\in I\). By the cancellation law for groups, \(h_i\) is unique. Thus there is a unique \(h\in\bigcap_i H_i\) such that \(ha=m\), and \(ha\in(\bigcap_i H_i)a\). Therefore, \((\bigcap_i H_i)a=\bigcap_i H_ia\).\(\quad\square\)
2. \(\quad\)(a) Let \(H\) be the cyclic subgroup (of order \(2\)) of \(S_3\) generated by \(\begin{pmatrix} 1&2&3\\ 2&1&3\\ \end{pmatrix}\). Then no left coset of \(H\) (except \(H\) itself) is also a right coset. There exists \(a\in S_3\) such that \(aH\cap Ha= \{a\}\).
\(\quad\)(b) If \(K\) is the cyclic subgroup (of order \(3\)) of \(S_3\) generated by \(\begin{pmatrix} 1&2&3\\ 2&3&1\\ \end{pmatrix}\), then every left coset of \(K\) is also a right coset of \(K\).
Proof. \(\quad\)It is easy to show them.\(\quad\square\)
3. The following conditions on a finite group \(G\) are equivalent.
\(\quad\)(i) \(|G|\) is prime.
\(\quad\)(ii) \(G\neq\langle e\rangle\) and \(G\) has no proper subgroups.
\(\quad\)(iii) \(G\cong Z_p\) for some prime \(p\).
Proof. \(\quad\)Case (i)\(\Rightarrow\)(ii). Let \(H\) be a subgroup of \(G\). By Corollary 4.6, the order of \(H\) divides the order of \(G\). So if \(|G|\) is prime, then \(|H|\) is either \(|G|\) or \(1\); thus \(G\) has no proper subgroups, and also clearly \(G\neq\langle e\rangle\). Case (ii)\(\Rightarrow\)(iii). If \(G\neq\langle e\rangle\) and \(G\) has no proper subgroups, then every element of \(G\) but \(e\) is a generator of \(G\); thus \(G\) is cyclic and \(|G|>1\), and so \(G\cong Z_p\) for some positive \(p>1\). If \(p>1\) is not prime, then \(Z_p\) has a proper subgroup; a contradiction. Case (iii)\(\Rightarrow\)(i). Clear.\(\quad\square\)
4. (Euler-Fermat) Let \(a\) be an integer and \(p\) a prime such that \(p\not\mid a\). Then \(a^{p-1}\equiv 1\) (mod \(p\)). [Hint: Consider \(\overline{a}\in Z_p\) and the multiplicative group of nonzero elements of \(Z_p\); see Exercise 1.7.] It follows that \(a^p\equiv a\) (mod \(p\)) for any integer \(a\).
Proof. \(\quad\)Recall that \(G=Z_p - \{0\}\) forms a multiplicative group of order \(p-1\). Let \(\overline{a}\in G\). By Lagrange’s Theorem, \(|\overline{a}|\) divides \(p-1\). Therefore, \(a^{p-1}\equiv 1\) (mod \(p\)) and \(a^p\equiv a\) (mod \(p\)).\(\quad\square\)
5. Prove that there are only two distinct groups of order \(4\) (up to isomorphism), namely \(Z_4\) and \(Z_2\bigoplus Z_2\). [Hint: By Lagrange’s Theorem 4.6 a group of order \(4\) that is not cyclic must consist of an identity and three elements of order \(2\).]
Proof. \(\quad\)Recall that every cyclic group of order \(4\) is isomorphic to \(Z_4\). So we assume that \(G\) is not cyclic. Then \(G\) consists of one element of order \(1\), i.e., identity and three elements of order \(2\); otherwise, at least one element of \(G - \{e\}\) has order \(4\), and this implies that \(G\) is cyclic, a contradiction. If \(a,b\in G\), then \((ab)^2=e\) and \(a^2b^2=ee=e\); thus \((ab)^2=a^2b^2\), and so \(G\) is abelian. Let \(G=\{e, a, b, c\}\). Since \(b^2=e\), \(b=b^{-1}\); thus \(ab\neq e\), and since \(a\neq e\) and \(b\neq e\), \(ab\neq a\) and \(ab\neq b\), and so \(ab=c\). Since \(G\) is abelian, \(ba=c\). Similarly, \(ac=ca=b\) and \(bc=bc=a\). Define \(f:Z_2\bigoplus Z_2\to G\) as follows: \[ f(0,0) = e,f(1,0) = a,f(0,1) = b,f(1,1) = c. \] Now we show that \(f\) is a homomorphism. \[ f((0,0)+(x,y))=f((x,y))=ef((x,y))=f((0,0))f((x,y)),\\ f((0,1)+(0,1))=e=b^2=f((0,1))f((0,1)),\\ f((0,1)+(1,0))=c=ba=f((0,1))f((1,0)),\\ f((0,1)+(1,1))=a=bc=f((0,1))f((1,1)),\\ f((1,1)+(1,1))=e=c^2=f((1,1))f((1,1)),\\ f((1,0)+(1,0))=e=a^2=f((1,0))f((1,0)),\\ f((1,1)+(1,0))=b=ca=f((1,1))f((1,0)). \] Since \(G\) is abelian, above cases are enough to show that \(f\) is a homomorphism. Since \(f\) is bijective, \(f\) is a isomorphism; thus \(G\cong Z_2\bigoplus Z_2\). Therefore, there are only two distinct groups of order 4.\(\quad\square\)
6. Let \(H,K\) be subgroups of a group \(G\). Then \(HK\) is a subgroup of \(G\) if and only if \(HK=KH\).
Proof. \(\quad\)Case \(\Rightarrow\). It is easy to show that if \(G\) is a group then \(G=\{g^{-1}\mid g\in G\}\). Since \(H\) and \(K\) are groups, \(HK=\{hk\mid h\in H, k\in K\}=\) \(\{h^{-1}k^{-1}\mid h\in H, k\in K\}\). Since \(HK\) is a group, \(HK=\{(h^{-1}k^{-1})^{-1}\mid h\in H, k\in K\}=\) \(\{kh\mid h\in H, k\in K\}=KH\).
\(\quad\)Case \(\Leftarrow\). Clearly, \(e\in HK\). Let \(x,y\in HK\). \(x=hk\) and \(y=h'k'\) for some \(h,h'\in H\) and \(k,k'\in K\); thus \(xy=hkh'k'\). Since \(HK=KH\), \(kh'=h''k''\) for some \(h''\in H\) and \(k''\in K\); thus \(xy=hh''k''k'\in HK\). \(x^{-1}=k^{-1}h^{-1}\). Similarly, \(k^{-1}h^{-1}=h'''k'''\) for some \(h'''\in H\) and \(k'''\in K\); thus \(x^{-1}\in HK\). Therefore, \(HK\) is a group.\(\quad\square\)
7. Let \(G\) be a group of order \(p^km\), with \(p\) prime and \((p,m) = 1\). Let \(H\) be a subgroup of order \(p^k\) and \(K\) a subgroup of order \(p^d\), with \(0<d<k\) and \(K\not\subset H\). Show that \(HK\) is not a subgroup of \(G\).
Proof. \(\quad\)By Theorem 4.7, \(|HK|=|H||K|/|H\cap K|\). Since \(H\cap K\subset K\), \(|H\cap K|=p^q\) for some \(0<q<d\); thus \(|HK|=p^{k+d-q}\). Since \(k+d-q>k\) and \((p,m)=1\), \(|HK|\) does not divides \(|G|\). Therefore, \(HK\) is not a subgroup of \(G\).\(\quad\square\)
8. If \(H\) and \(K\) are subgroups of finite index of a group \(G\) such that \([G:H]\) and \([G:K]\) are relatively prime, then \(G=HK\).
Proof. \(\quad\)Notice that \([G:K][K:H\cap K]=\) \([G:H][H:H\cap K]=\) \([G:H\cap K]\). Since \([G:H]\) and \([G:K]\) are relatively prime, \([G:K]\mid[H:H\cap K]\). By Proposition 4.8, \([H : H \cap K] \le [G : K]\). Thus \([G:K]=[H:H\cap K]\), and so \([G:H\cap K]=[G:H][G:K]\). Therefore, also by Proposition 4.8, \(G=HK\).\(\quad\square\)
9. If \(H,K\) and \(N\) are subgroups of a group \(G\) such that \(H<N\), then \(HK\cap N=H(K\cap N)\).
Proof. \(\quad\)Notice that \(H(K\cap N) = \{hx\mid h\in H, x\in K\cap N\}=\) \(\{hk=hn\mid h\in H, k\in K, n\in N\}=\) \(HK \cap HN\). Since \(H<N\) and \(e\in H\), \(HN=N\). Therefore, \(HK\cap N=HK\cap HN=H(K\cap N)\).\(\quad\square\)
10. Let \(H,K,N\) be subgroups of a group \(G\) such that \(H<K, H\cap N=K\cap N\), and \(HN=KN\). Show that \(H=K\).
Proof. \(\quad\)\(K\subset KN=HN\). Thus \(K\subset HN\cap K\). By Excercise 9, \(HN\cap K=H(K\cap N)=H(H\cap N)=H\). Thus \(K\subset H\). Therefore \(K=H\).\(\quad\square\)
11. Let \(G\) be a group of order \(2n\); then \(G\) contains an element of order \(2\). If \(n\) is odd and \(G\) abelian, there is only one element of order \(2\).
Proof. \(\quad\)By Exercise 1.14, if \(G\) is even, then \(G\) contains an element of order \(2\). Now suppose that \(n\) is odd, \(G\) is abelian, and \(a,b\in G\) such that \(a\neq b\) are of order \(2\). Consider \(S=\{e, a, b, ab\}\). It is easy to show that \(S\) forms a subgroup of \(G\), but since \(n\) is odd, \(4\not\mid 2n\); a contradiction.\(\quad\square\)
12. If \(H\) and \(K\) are subgroups of a group \(G\), then \([H\vee K:H]\ge[K: H\cap K]\).
Proof. \(\quad\)By Proposition 4.8, obvious; \(H\) and \(K\) are subgroups of \(H\vee K\).\(\quad\square\)
13. If \(p>q\) are primes, a group of order \(pq\) has at most one subgroup of order \(p\). [Hint: Suppose \(H,K\) are distinct subgroups of order \(p\). Show \(H\cap K=\langle e\rangle\); use Exercise 12 to get a contradiction.]
Proof. \(\quad\)Since \(H\cap K\) is a subgroup of \(H\) and \(p\) is prime, \(H\cap K\) is \(\langle e\rangle\) or \(H\). Since \(H,K\) are distinct, \(H\cap K\) is \(\langle e\rangle\). By Exercise 12, \([K: H\cap K]=|K|=p<\) \([H\vee K:H]\); thus \(p^2\le|H\vee K|\), and so \(|H\vee K|>pq\); a contradiction.\(\quad\square\)
14. Let \(G\) be a group and \(a,b\in G\) such that (i) \(|a|=4=|b|\); (ii) \(a^2=b^2\); (iii) \(ba=a^3b=a^{-1}b\); (iv) \(a\neq b\); (v) \(G=\langle a,b\rangle\). Show that \(|G|=8\) and \(\cong Q_8\). (See Exercise 2.3; observe that the generators \(A,B\) of \(Q_8\) also satisfy (i)-(v).)
Proof. \(\quad\)Consider \(S=\{e,a,a^2,a^3,b,ab,a^2b,a^3b\}\). It is easy to show that \(S\) is the only group satisfying the properties above.\(\quad\square\)