A solutions manual for Algebra by Thomas W. Hungerford
Chapter I: Groups
1. Semigroups, Monoids, and Groups
1. Give examples other than those in the text of semigroups and monoids that are not groups.
Example. \(\quad\)A power set of a set with union forms an abelian monoid that is not a group. The positive integers under addition form a semigroup that is not a monoid.
2. Let \(G\) be a group (written additively), \(S\) a nonempty set, and \(M (S,G)\) the set of all functions \(f : S\to G\). Define additions in \(M (S, G)\) as follows: \((f + g) : S\to G\) is given by \(s\mapsto f(s)+g(s)\in G\). Prove that \(M(S, G)\) is a group, which is abelian if \(G\) is.
Proof. \(\quad\)Let \(s\in S\) and let \(f, g\), and \(h\in M(S, G)\). Since \((f+g)(x)\) has a unique value \(f(x)+g(x)\in G\), \(f+g\) is a well-defined function in \(M(S, G)\). \(((f+g)+h)(x)\) \(=\) \((f+g)(x)+h(x)\) \(=\) \((f(x)+g(x))+h(x)\) \(=\) \(f(x)+(g(x)+h(x))\) \(=\) \(f(x)+(g+h)(x)\) \(=\) \((f+(g+h))(x)\); thus associative. Similarly, if \(G\) is abelian, then \(f+g=g+f\). Let \(0_S\) be a function in \(M(S,G)\) given by \(x\mapsto e\), and let \(-f\) be a function in \(M(S,G)\) given by \(x\mapsto f(x)^{-1}\). It is clear that \(0_S\) is an identity element and \(-f\) is an inverse element of \(f\in M(S, G)\).\(\quad\square\)
3. Is it true that a semigroup which has a left identity element and in which every element has a right inverse (see Proposition 1.3) is a group?
Proof. \(\quad\)Let \(S\) be a set such that \(|S|>1\), and define \(xy=y\) for all \(x,y\in S\). \((xy)z=x(yz)=z\) for all \(x,y,z\in S\); thus a semigroup. Pick any \(a\in S\). Since \(ax=x\) for all \(x\in S\), \(a\) is a left identity element, and also since \(xa=a\) for all \(x\), every \(x\) has a right inverse \(a\). But for \(x\) such that \(a\neq x\in S\), \(x=ax\neq xa=a\); thus not a group.\(\quad\square\)
4. Write out a multiplication table for \(D_4^\ast\).
Example. \(\quad\) \[ \begin{array}{l|llllllll} & I & R & R^2 & R^3 & T_x & T_{2,4} & T_y & T_{1,3}\\ \hline I & I & R & R^2 & R^3 & T_x & T_{2,4} & T_y & T_{1,3}\\ R & R & R^2 & R^3 & I & T_{2,4} & T_y & T_{1,3} & T_x\\ R^2 & R^2 & R^3 & I & R & T_y & T_{1,3} & T_x & T_{2,4}\\ R^3 & R^3 & I & R & R^2 & T_{1,3} & T_x & T_{2,4} & T_y\\ T_x & T_x & T_{2,4} & T_y & T_{1,3} & I & R^3 & R^2 & R\\ T_{2,4} & T_{2,4} & T_x & T_{1,3} & T_y & R & I & R^3 & R^2\\ T_y & T_y & T_{1,3} & T_x & T_{2,4} & R^2 & R & I & R^3\\ T_{1,3} & T_{1,3} & T_y & T_{2,4} & T_x & R^3 & R^2 & R & I\\ \end{array} \]
5. Prove that the symmetric group on \(n\) letters, \(S_n\), has order \(n!\).
Proof. \(\quad\)\(S_1=1\), \(|S_n|=n|S_{n-1}|=n!\).\(\quad\square\)
6. Write out an addition table for \(Z_2\bigoplus Z_2\). \(Z_2\bigoplus Z_2\) is called the Klein Four Group.
Example. \(\quad\) \[ \begin{array}{c|cccc} & (0,0) & (0,1) & (1,0) & (1,1)\\ \hline (0,0)& (0,0) & (0,1) & (1,0) & (1,1)\\ (0,1)& (0,1) & (0,0) & (1,1) & (1,0)\\ (1,0)& (1,0) & (1,1) & (0,0) & (0,1)\\ (1,1)& (1,1) & (1,0) & (0,1) & (0,0)\\ \end{array} \]
7. If \(p\) is prime, then the nonzero elements of \({Z}_p\) form a group of order \(p - 1\) under multiplication. [Hint: \(\overline{a}\ne \overline{0}\Rightarrow (a, p) = 1\); use Introduction, Theorem 6.5.] Show that this statement is false if p is not prime.
Proof. \(\quad\)Let \(a_1, a_2, b_1\), and \(b_2\) be integers. If \(\overline{a}_1=\overline{a}_2\) and \(\overline{b}_1=\overline{b}_2\), then \(a_1 = a_2+mp\) and \(b_1 = b_2+np\) for some \(m,n\in\mathbb{Z}\). Thus \(a_1b_1\) \(=\) \((a_2+mp)(b_2+np)\) \(=\) \(a_2b_2+a_2np+mpb_2+mpnp\) \(\sim\) \(a_2b_2\), and so congruence modulo \(p\) is a congruence relation respect to multiplication. Since \(\mathbb{Z}\) is a commutative monoid under multiplication, \(Z_p\) forms a commutative monoid under multiplication. If \(\overline{a}\ne \overline{0}\) and \(\overline{b}\ne \overline{0}\), then \(p\nmid ab\); otherwise \(p\mid a\) or \(p\mid b\), a contradiction. Thus if we let \(Z^\ast_p=\{a\in Z_p|a\neq\overline{0}\}\), then \(Z^\ast_p\) is closed under multiplication, and also has an identity element \(\overline{1}\). Let \(a\in \bigcup Z^\ast_p\). Since \(Z^\ast_p\) is finite, \(a^m=a^n\) for some \(0<m<n\), \(a^m=a^ma^{n-m}\) \(\Leftrightarrow\) \(a^{n-m}\in\overline{1}\) \(\Leftrightarrow\) \(a^{n-m-1}a\) \(=\) \(aa^{n-m-1}\). Since \(a\neq 0\), \(\overline{a^{n-m-1}}\in Z^\ast_p\) is an inverse element of each \(\overline{a}\). Thus \(Z^\ast_p\) forms a group under multiplication. Suppose that \(p\) is not prime. Let \(p=4\). Since there is no inverse element of \(\overline{0}\), we only show that \(\{\overline{1}, \overline{2}, \overline{3}\}\) does not form a group. \(\overline{2}\overline{2}=\overline{0}\), thus \(\{\overline{1}, \overline{2}, \overline{3}\}\) is not closed under multiplication.\(\quad\square\)
8. \(\quad\)(a) The relation given by \(a\sim b\Leftrightarrow a-b\in Z\) is a congruence relation on the additive group \(\mathbb{Q}\) [see Theorem 1.5].
\(\quad\)(b) The set \(\mathbb{Q}/\mathbb{Z}\) of equivalence classes is an infinite abelian group.
Proof. \(\quad\)(a) It is easily seen that \(\sim\) is reflexive, symmetric, and transitive; thus an equivalence relation on \(\mathbb{Q}\). Let \(a_1, a_2, b_1, b_2\in\mathbb{Q}\). If \(\overline{a}_1=\overline{a}_2\) and \(\overline{b}_1=\overline{b}_2\), then \(a_1-a_2=m\) and \(b_1-b_2=n\) for some \(m\) and \(n\in\mathbb{Z}\). \((a_1+b_1)-(b_2+b_2)\) \(=\) \(m+n\). Thus \(a_1+b_1\sim a_2+b_2\).
\(\quad\)(b) Let \(m\) and \(n\) be positive integers, and let \(k\) be an integer. By Theorem 1.5, the set forms an abelian group. If \(1/m=1/n+k\), then \(k=0\), thus \(m=n\), and so \(\{\overline{1/{n}}\in\mathbb{Q}/\mathbb{Z}\mid n\in\mathbb{N}^\ast\}\) is countable.\(\quad\square\)
9. Let \(p\) be a fixed prime. Let \(R_p\) be the set of all those rational numbers whose denominator is relatively prime to \(p\). Let \(R^p\) be the set of rationals whose denominator is a power of \(p\) (\(p^i,i\ge 0\)). Prove that both \(R_p\) and \(R^p\) are abelian groups under ordinary addition of rationals.
Proof. \(\quad\)Let \(x,y,z,a/b\), and \(c/d\in R_p\). Since \(a/b+c/d=(ad+bc)/bd\) and \(bd\) is relatively prime to \(p\), \(a/b+c/d\in R_p\). Thus \(R_p\) is closed under ordinary addition. \((x+y)+z=x+(y+z)\). \(0=0/1\in R_p\), and \(x+0=0+x=x\). \(-a/b\in R_p\) for each \(a/b\in R_p\) and \(a/b+(-a/b)=0\). \(x+y=y+x\).
\(\quad\)Let \(a/b\) and \(c/d\in R^p\). Since \(a/b+c/d=(ad+bc)/bd\), \(bd=p^i\) for some \(i\in\mathbb{N}\), and if \(bd\) is reduced to \(b'd'\), then also \(b'd'=p^j\) for some \(j\in\mathbb{N}\). Thus \(R^p\) is closed under ordinary addition. Similarly to the previous, \(R^p\) is abelian groups under ordinary addition of rationals.\(\quad\square\)
10. Let \(p\) be a prime and let \(Z(p^\infty )\) be the following subset of the group \(\mathbb{Q}/\mathbb{Z}\) (see pg. 27): \[ Z(p^\infty )=\{\overline{a/b}\in\mathbb{Q}/\mathbb{Z}\mid a,b\in Z\text{ and } b=p^i\text{ for some } i\ge 0\}. \] Show that \({Z}(p^\infty )\) is an infinite group under the addition operation of \(\mathbb{Q}/\mathbb{Z}\).
Proof. \(\quad\)Let \(a/b\) and \(c/d\in \mathbb{Q}\). If \(a/b\sim c/d\) and \(b=p^i\) for some \(i\ge 0\), then clearly, \(d=p^j\) for some \(j\ge 0\). Thus \(Z(p^\infty )\) is a subset of the group \(\mathbb{Q}/\mathbb{Z}\). It is easily seen that \(Z(p^\infty)\) is closed under the addition operation of \(\mathbb{Q}/\mathbb{Z}\), and has an identity element \(\overline{0}=\overline{0/p}\), and that there is an inverse element \(-\overline{a/b}\) of each \(\overline{a/b}\). Similarly to exercise 8 (b), \(\{\overline{1/p^n}\in Z(p^\infty)\mid n\in\mathbb{N}^\ast\}\) is countable.\(\quad\square\)
11. The following conditions on a group \(G\) are equivalent: (i) \(G\) is abelian; (ii) \((ab)^2 = a^2b^2\) for all \(a,b\in G\); (iii) \((ab)^{-1} = a^{-1}b^{-1}\) for all \(a,b\in G\); (iv) \((ab)^n = a^nb^n\) for all \(n\in \mathbb{Z}\) and all \(a,b\in G\); (v) \((ab)^n = a^nb^n\) for three consecutive integers \(n\) and all \(a, b\in G\). Show (v) \(\Rightarrow\) (i) is false if “three” is replaced by “two.”
Proof. \(\quad\)Case (i)\(\Rightarrow\)(ii). \((ab)^2=abab=aabb=a^2b^2\).
\(\quad\)Case (ii)\(\Rightarrow\)(i). \(abab=aabb\) \(\Leftrightarrow\) \(a^{-1}ababb^{-1}=a^{-1}aabbb^{-1}\) \(\Leftrightarrow\) \(ba=ab\).
\(\quad\)Case (i)\(\Rightarrow\)(iii). \((a^{-1}b^{-1})ab=a^{-1}b^{-1}ba=a^{-1}a=e\). \(ab(a^{-1}b^{-1})=aba^{-1}b^{-1}=aa^{-1}bb^{-1}=e\).
\(\quad\)Case (iii)\(\Rightarrow\)(i). \((ab)^{-1}=a^{-1}b^{-1}\) \(\Leftrightarrow\) \((ab)^{-1}ab=a^{-1}b^{-1}ab\) \(\Leftrightarrow\) \(e=a^{-1}b^{-1}ab\) \(\Leftrightarrow\) \(a=b^{-1}ab\) \(\Leftrightarrow\) \(ba=ab\).
\(\quad\)Case (i), (iii)\(\Rightarrow\)(iv). Let \(P(n)\) be the property “\((ab)^n = a^nb^n\)”. \(P(0)\) holds. If \(P(n)\) holds, then \((ab)^{n+1}=a^nb^nab=a^nab^nb=a^{n+1}b^{n+1}\), thus \(P(n+1)\) holds. Let \(Q(n)\) be the property “\((ab)^{-n} = a^{-n}b^{-n}\)”. \(Q(1)\) holds by (iii). If \(Q(n)\) holds, then \((ab)^{-(n+1)}=a^{-n}b^{-n}(ab)^{-1}=a^{-(n+1)}b^{-(n+1)}\), thus \(Q(n+1)\) holds. Therefore, \((ab)^n = a^nb^n\) for all \(n\in \mathbb{Z}\) and all \(a,b\in G\).
\(\quad\)Case (iv)\(\Rightarrow\)(ii). Clear.
\(\quad\)Case (iv)\(\Rightarrow\)(v). Clear.
\(\quad\)Case (v)\(\Rightarrow\)(i). Let \(a\) and \(b\in G\) and let \(n\in\mathbb{Z}\). \((ab)^n = a^nb^n, (ab)^{n+1}=a^{n+1}b^{n+1}, (ab)^{n+2}=a^{n+2}b^{n+2}\). \(a^{n+1}b^{n+1}\) \(=\) \((ab)^{n+1}\) \(=\) \((ab)^n(ab)\) \(=\) \(a^nb^nab\) \(\Leftrightarrow\) \(a^{-n}a^{n+1}b^{n+1}b^{-1}=a^{-n}a^nb^nabb^{-1}\) \(\Leftrightarrow\) \(ab^n=b^na\). By replacing \(a\) and \(b\) with \(b\) and \(a\), \(ba^n=a^nb\). \(a^{n+2}b^{n+2}=(ab)(ab)^{n+1}\) \(=\) \(aba^{n+1}b^{n+1}\) \(\Leftrightarrow\) \(a^{n+2}bb^{n+1}=aba^{n+1}b^{n+1}\) \(\Leftrightarrow\) \(aa^{n+1}b=aba^{n+1}\) \(\Leftrightarrow\) \(a^{n+1}b=ba^{n+1}\). Similarly, \(ab^{n+1}=b^{n+1}a\). Finally, \(b(ab^n)=b(b^na)=b^{n+1}a=ab^{n+1}\) \(\Leftrightarrow\) \(bab^n=abb^n\) \(\Leftrightarrow\) \(ab=ba\).
\(\quad\)Case (v)\(\Rightarrow\)(i). \((ab)^n = a^nb^n\) holds for \(n=0\) and \(n=1\) in all groups. Thus (v) \(\Rightarrow\) (i) is false.\(\quad\square\)
12. If \(G\) is a group, \(a,b\in G\) and \(bab^{-1}=a^{r}\) for some \(r\in\mathbb{N}\), then \(b^jab^{-j}=a^{r^j}\) for all \(j\in\mathbb{N}\).
Proof. \(\quad\)Let \(P(n)\) be the property “\(b^nab^{-n}=a^{r^n}\)”. \(a^{r^0}=a=b^0ab^{-0}=a\). \(P(0)\) holds. If \(P(n)\) holds, then \(a^{r^{n+1}}=a^{r^1r^{n}}=(a^{r})^{r^{n}} =(bab^{-1})(bab^{-1})\cdots bab^{-1}\) \(=\) \(ba^{r^{n}}b^{-1}\) \(=\) \(bb^nab^{-n}b^{-1}\) \(=\) \(b^{n+1}ab^{-(n+1)}\). \(P(n+1)\) holds.\(\quad\square\)
13. If \(a^2 = e\) for all elements \(a\) of a group \(G\), then \(G\) is abelian.
Proof. \(\quad\)If \(a,b\in G\), then \((ab)^2=e=a^2b^2\). By exercise 11 (ii), \(G\) is abelian.\(\quad\square\)
14. If \(G\) is a finite group of even order, then \(G\) contains an element \(a\ne e\) such that \(a^2 = e\).
Proof. \(\quad\)Let \(a\in G\). Notice that if \(a^2=e\) then \(a=a^{-1}\). Define the relation \(a\sim b\) if and only if \(a=b\) or \(a=b^{-1}\). It is easily seen that \(\sim\) is an equivalence relation on \(G\). Notice that for each \(a\in G\), the order of \(\overline{a}\) is at most \(2\), and that if \(a\neq a^{-1}\) then \(|\overline{a}|=2\). Now consider \(S=\{\overline{a}\mid a\neq e, a\in G\}\). If \(G\) is finite and there is no element \(a\neq e\) such that \(a^2=e\), i.e., \(a=a^{-1}\), then \(\bigcup S=G \smallsetminus\{e\}\) has an even order; thus \(G\) has an odd order. Therefore, if \(G\) is finite and \(G\) has an even order, then \(G\) contains an element \(a\ne e\) such that \(a^2 = e\).\(\quad\square\)
15. Let \(G\) be a nonempty finite set with an associative binary operation such that for all \(a,b,c\in G\text{ } ab=ac\Rightarrow b=c\text{ and } ba=ca\Rightarrow b=c\). Then \(G\) is a group. Show that this conclusion may be false if \(G\) is infinite.
Proof. \(\quad\)Let \(a,b\in G\). Since \(G\) is finite and closed under the operation, \(a^m=a^n\) for some \(m\) and \(n\) such that \(1<m<n\). Since the operation is associative, \(aa^{m-1}=aa^{n-m+1}a^{m-1}\), and using the right cancelation law, \(a=aa^{n-m+1}\). Similarly, \(a=a^{n-m+1}a\). Thus \(a=a^{k_a}a=aa^{k_a}\) for some \(k_a>1\). \(ab=aa^{k_a}b\), thus \(b=a^{k_a}b\). \(ba=ba^{k_a}a\), thus \(b=ba^{k_a}\). \(a^{k_a}\) is an identity element of \(G\), and similarly \(a=ab^{k_b}=aa^{k_a}\), thus \(a^{k_a}=b^{k_b}\). Since \(a^{k_a}=aa^{k_a-1}=a^{k_a-1}a\), there is an inverse element of each \(a\in G\).
\(\quad\)The positive integers under addition form a semigroup that is not a monoid; the left and the right cancelation laws hold for the semigroup.\(\quad\square\)
16. Let \(a_1, a_2,...\) be a sequence of elements in a semigroup \(G\). Then there exists a unique function \(\psi :\mathbb{N}^\ast\to G\) such that \(\psi (1) = a_1,\psi (2) = a_1a_2,\psi (3) = (a_1a_2)a_3\) and for \(n\ge 1, \psi (n+1) = (\psi (n))a_{n+1}\). Note that \(\psi (n)\) is precisely the standard \(n\) product \(\prod_{i=1}^n{a_i}\) [Hint: Applying the Recursion Theorem 6.2 of the Introduction with \(a=a_1,S=G\) and \(f_n :G\to G\) given by \(x\to xa_{n+2}\) yields a function \(\varphi :\mathbb{N}\to G\). Let \(\psi =\varphi\theta\), where \(\theta :\mathbb{N}^\ast\to\mathbb{N}\) is given by \(k\mapsto k-1\).]
Proof. \(\quad\)\(f_n\) is well-defined for each \(n\in\mathbb{N}\). By the Recursion Theorem 6.2 of the Introduction, we have a unique function \(\varphi:\mathbb{N}\to G\) such that \(\varphi(0)=a_1\) and \(\varphi(n+1)=f_n(\varphi(n))\) for every \(n\in\mathbb{N}\). \(\varphi(0)=a_1,\varphi(1)=a_1a_2, \varphi(2)=(a_1a_2)a_3\) and for \(n\ge 0\), \(\varphi(n+1)=(\varphi(n))a_{n+1}\). Finally, we have a unique function \(\psi =\varphi\theta\) such that \(\psi (1) = a_1,\psi (2) = a_1a_2,\psi (3) = (a_1a_2)a_3\) and for \(n\ge 1, \psi (n+1) = (\psi (n))a_{n+1}\).\(\quad\square\)