202101141527 Homework 2 (Q3)

Let $G$ be a group and $e\neq a\in G$ in which $e$ is the identity of $G$ . Suppose $\mathrm{ord}(a)=n$ .

i. If $a^h=e$ , show that $n\big| h$ .

ii. Evaluate the order of $a^m$ where $m$ is an integer.

iii. Show that $|H|=n/(m,n)$ if $G=\langle a\rangle$ and $H=\langle a^m\rangle$ where $m\in\mathbb{N}$ .

iv. If G is cyclic of order $n$ , show that for any $d\big| n$ , there exists a subgroup of order $d$ in $G$ . Could $G$ have more than one subgroup of order $d$ ? (Clearly the answer is no if $d=1$ or $n$ .) Justify with a proof (if you think yes) or give a counterexample (if you think no).

Attempts.

i. By Division Algorithm, $h=qn+r$ for some $q\in\mathbb{Z}$ and $0\le r<n$ . Then one may write $a^r=a^{h-qn}=a^h(a^n)^{-q}=e$ . It can be seen that $r$ must be $0$ lest $a^r=e$ where $1\le r<n$ contradicts to the assumption that $n=\mathrm{ord}(a)$ where $n$ is the smallest positive integer such that $a^n=e$ . Thus $h=qn$ , and $n|h$ .

ii. First, $(a^m)^{\frac{n}{(n,m)}}=a^{\frac{mn}{(n,m)}}=(a^n)^{\frac{m}{(n,m)}}=e^{\frac{m}{(n,m)}}=e$ . Secondly, if $k\in\mathbb{N}$ such that $(a^m)^k=e$ , from the result of part i. previously, it follows that $\frac{n}{(n,m)}|k$ because if $a|bc$ and $(a,b)=1$ , then $a|c$ (here $a=\frac{n}{(n,m)}$ ), $b=\frac{m}{(n,m)}$ , and $c=k$ . Such a $k$ will always be greater than or equal to $\frac{n}{(n,m)}$ . Thus $\mathrm{ord}(a^m)=\frac{n}{(n,m)}$ .

iii. If $H=\langle a^m\rangle$ then $|H|=\mathrm{ord}(a^m)\stackrel{\textrm{by (ii)}}{=}\frac{n}{(n,m)}$ .

iv. Let $G=\langle a\rangle$ as it is cyclic. Let $m=n/d$ where $d|n$ is given in the problem. The order of the cyclic subgroup $\langle a^m\rangle$ of $G$ is then $\frac{n}{(n,m)}=\frac{n}{m}=d$ , in view of the result in part iii. We have constructed a subgroup of order $d$ in $\langle a\rangle$ where $d|n$ .

We wish to prove that it is unique: If $H$ is a subgroup of $G=\langle a\rangle$ and $|H|=d$ , then $H=\langle a^{\frac{n}{d}}\rangle$ . Note that $H=\langle a^m\rangle$ and $\frac{n}{(n,m)}=d$ . Consider the subgroup $\langle a^{\frac{n}{d}}\rangle$ . Because $\frac{n}{d}=(n,m)$ divides $m$ , we have $a^m\in \langle a^{\frac{n}{d}}\rangle$ and thus $H\subset \langle a^{\frac{n}{d}}\rangle$ . From the result of part iii. it follows that $|\langle a^{\frac{n}{d}}\rangle | =\frac{n}{(n,\frac{n}{d})}=d$ . Thus $H=\langle a^{\frac{n}{d}}\rangle$ is unique.

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