【Basic Abstract Algebra】Exercises for Section 2.3 — Cyclic groups

1. Let \(G\) be a group, \(a,b\in G\), and \(ab=ba,~|a|=m,~|b|=n,~\gcd(m,n)=1\). Show that \(|ab|=mn\).
   Proof: Since \(|a|=m,~|b|=n\), we have \(a^m=b^n=e\), where \(e\) is the identity of \(G\). Because \(\gcd(m,n)=1\), so \(\text{lcm}(m,n)=mn\). For any \(k\in\mathbb Z\), we have \((ab)^k=a^kb^k\) since \(ab=ba\). Note that \((ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e\), then \(|ab|\mid mn\), it follows that \(|ab|=mn\), or (\(|ab|\mid m\) and \(|ab|\not\mid n\)) or (\(|ab|\not\mid m\) and \(|ab|\mid n\)) since \(\gcd(m,n)=1\). Suppose that \(|ab|\mid m\) and \(|ab|\not\mid n\), then \((ab)^m=a^mb^m=b^m=1\Rightarrow |b|\mid m\Rightarrow n\mid m~\to\leftarrow\). Similarly, \(|ab|\not\mid m\) and \(|ab|\mid n\Rightarrow ~\to\leftarrow\). Thus, \(|ab|=mn\). #

2. Show that the group with prime order is a cyclic group.
   Proof: Let \(G\) be a group, \(p\) be a prime, and \(|G|=p\), which means \(G\) has \(p\) elements. Let \(g\in G\) and \(g\neq e\) (identity). Since \(|G|=p\), then \(|g|\mid p\). Therefore, \(|g|=1\) or \(p\). If \(|g|=1\), then \(g=e~\to\leftarrow\). Thus, \(|G|=p\), and \(\langle g\rangle=\{e,g,g^2,\cdots,g^{p-1}\}\) has \(p\) elements. Because \(G\) has also \(p\) elements and \(\langle g\rang\le G\), we have \(\lang g\rang=G\), i.e. \(G\) is a cyclic group. #

3. Let \(G\) be a group, \(g\) in \(G\), \(|g|=mn\), and \(\gcd(m,n)=1\), then \(g=ab\) where \(|a|=m\), \(|b|=n\), and \(a,b\in G\).
   Proof: Since \(\gcd(m,n)=1\), then \(\exists ~u,v\in\mathbb Z,~s.t.~um+vn=1\). Let \(a = g^{vn},~b=g^{um}\), then

   \[ab=g^{vn}\cdot g^{um}=g^{vn+um}=g^1=g \]

   We want to show that \(|a|=m,~|b|=n\). Since \(|g|=mn\), then \(g^{mn}=e\) (identity). Thus, \(a^m={g^{vn}}^m=g^{mn\cdot v}=e^v=e\), so \(|a|\mid m\). Let \(0<k< m\), if \(a^k=a^{kvn}=e\), then \(mn\mid kvn\Rightarrow m\mid kv\Rightarrow m\mid v\). Then \(\exists~x\in\mathbb Z,~s.t. ~v=xm\) \(\Rightarrow m(u+xn)=1\Rightarrow m=1\), then we can not find a \(0<k<m\), such that \(a^k=e\). So \(|a|=m\). Similarly, \(|b|=n\). #

4. Let \(a,b\) be elements of group \(G\) such that \(a^3=b^2=e,~(ab)^2=e,~a^2\neq e,~b\neq e\). What is \(\lang a,b\rang\).
   Solution: Since \(a^3=b^2=e,~a^2\neq e,~b\neq e\), we have \(|a|=3,~|b|=2\). Then

   \[\lang a\rang=\{e,a,a^2\},\quad \lang b\rang=\{e,b\} \]

   If \(ab=e\), then \(ab\cdot b=b\Rightarrow a=b\Rightarrow a^2=b^2=e\), it is contradicts to the fact that \(a^2\neq e\). Therefore, \(ab\neq e\). Thus, \(|ab|=2\) since \((ab)^2=e\). Clearly, \(a\neq b\). And \(aba=(ab)^2b^{-1}=b^{-1}=b,~bab=a^{-1}(ab)^2=a^{-1}=a^2\). Therefore, \(\lang a,b\rang=\{e,a,b,a^2,ab,ba,a^2b\}\).

5. Let \(a,b\) be elements of a group \(G\) such that \(a^3=b^2=e,~ab=ba,~a^2\neq e,~b\neq e\). What is \(\lang a,b\rang\).
   Solution: \(|a|=3,~|b|=2\). Since \(ab=ba\), we have \(\lang a,b\rang=\{e,a,b,ab,a^2,a^2b\}\).

6. Let \(G\) be a finite abellian group. Prove that the product of all the elements of \(G\) equals the product of all the elements of \(G\) of order \(2\).
   Proof: Let \(g\in G\), and \(|g|\neq2\). If \(|g|=1\), then \(g=e\) (identity). Suppose that \(|g|>2\). Since \(G\) is a group, we can find \(g^{-1}\in G,~s.t.~ gg^{-1}=g^{-1}g=e\). We want to show that \(|g^{-1}|>2\). Suppose that \(|g^{-1}|=2\), i.e. \((g^{-1})^2=e\). Therefore,

   \[g^{-1}=g^{-1}\cdot e=g^{-1}\cdot(gg^{-1})=(g^{-1}g)g=g. \]

   It follows that \(|g|=2,~\to\leftarrow\). Thus, \(|g^{-1}|>2\). Which means that for any \(g\in G\) and \(|g|>2\), we can find a \(g^{-1}\in G,\) s.t. \(gg^{-1}=g^{-1}g=e\) and \(|g^{-1}|>2\). Thus,

   \[\prod_{g\in G}g=e\cdot\left(\prod_{g\in G~\&~|g|=2}g\right)\cdot\left(\prod_{g\in G~\&~|g|>2}g\right)=\left(\prod_{g\in G~\&~|g|=2}g\right)\cdot e=\left(\prod_{g\in G~\&~|g|=2}g\right). \]

7. Use the conclusion of the previous question to show the Wilson's Theorem: If \(p\) is a prime, then \((p-1)!=-1(\mod p)\).
   Proof: Consider the group \((\mathbb Z_p/\{\bar 0\},\times):=(G,\times)\), where \(G\) consists of all nonzero integers modulo \(p\), i.e., the set \(\{ \bar 1, \bar 2, \cdots, \overline{p-1} \}\). We now show that \(G\) is a abelian group.

   \(\times~(\mod p)\)	\(\bar 1\)	\(\bar2\)	\(\cdots\)	\(\overline{p-1}\)
   \(\bar 1\)	\(\bar1\)	\(\bar2\)	\(\cdots\)	\(\overline{p-1}\)
   \(\bar2\)	\(\bar2\)	\(\bar4\)	\(\cdots\)	\(\overline{p-2}\)
   \(\vdots\)	\(\vdots\)	\(\vdots\)		\(\vdots\)
   \(\overline{p-1}\)	\(\overline{p-1}\)	\(\overline{p-2}\)	\(\cdots\)	\(\bar1\)

   Clearly, for any \(\bar a,\bar b\in G\), we have \(\bar a\times \bar b \in G\).

   • For any \(\bar a,\bar b,\bar c\in G\), since \((a\times b)\times c=a\times(b\times c)\), we have \((\bar a\times \bar b)\times \bar c= \bar a\times(\bar b\times \bar c)\).
   • The identity of \(G\) is \(\bar1\) since \(\bar 1\times \bar a=\bar a\times \bar 1\) for any \(\bar a\in G\).
   • By the Fermat Little Theorem, for any \(\bar a\in G\), \(\bar a^{p-1}=\bar 1\), hence \(\bar a^{p-2}\) is the inverse of \(\bar a\).

   Therefore, \(G\) is a group. And \(G\) is an abelian group since \(\bar a\times \bar b=\bar b\times \bar a\) for any \(\bar a,\bar b\in G\). For any \(\bar x\in G\), let \(\bar x^2=1\), then there \(\exists n\in\mathbb Z\), s.t. \(x^2=np+1\). Note that \(x\in\{1,2,\cdots,p-1\}\), then \(0\le n\le p-2\). Let \(n=p-k,~2\le k\le p\), then \(np+1=p^2-kp+1\). \(x^2=p^2-kp+1\Rightarrow k^2-4=0\) or \(k=p\), i.e. \(k=2\) or \(k=p\). Thus \(x=p-1\) or \(x=1\). By the conclusion of the previous question, we have

   \[\bar 1\times \bar 2\times\cdots\times \overline{p-1}=\bar 1\times\overline{p-1}, \]

   i.e.

   \[(p-1)!\equiv p-1~(\mod p)\Rightarrow (p-1)!\equiv -1(\mod p).\quad \# \]