1. group element order

Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

Please show details of this proof. It is another practice problem for my final exam.

Thank you so much!

2. Originally Posted by mpryal
Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

Please show details of this proof. It is another practice problem for my final exam.

Thank you so much!
The first thing we should do is try to prove closure. Choose $\displaystyle a,b\in T$. Then there exists positive integers $\displaystyle m,n$ with $\displaystyle a^m=e$ and $\displaystyle b^n=e$, with $\displaystyle m,n>1$. Since $\displaystyle a,b\in G$, then $\displaystyle ab\in G$. But is it also in $\displaystyle T$? Well, observe that $\displaystyle (ab)^{mn}\in G$. Also, $\displaystyle (ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e^ne^m=e$. So then the conditions are satisfied such that $\displaystyle ab\in T$, and therefore $\displaystyle T$ is closed.

It is pretty obvious that $\displaystyle e\in T$.

It's almost just as obvious that for any $\displaystyle a\in T$ then $\displaystyle a^{-1}\in T$. But let's prove it anyway. There exists integer $\displaystyle m>1$ with $\displaystyle a^m=e$. Observe also that since $\displaystyle T$ is abelian, we can do the following: $\displaystyle e=a^{-1}a=(a^{-1}a)^m=(a^{-1})^ma^m=(a^{-1})^m$. So for any $\displaystyle a\in T$, then $\displaystyle a^{-1}\in T$.

That should do it.

3. Originally Posted by mpryal
Let G be an abelian group and let T={a in G|a^m=e for some m>1}. Prove that T is a subgroup of G and that G/T has no element - other than its identity element - of finite order.

Please show details of this proof. It is another practice problem for my final exam.

Thank you so much!
Hi mpryal.

Let $\displaystyle gT\in G/T$ and suppose $\displaystyle (gT)^k=e_{G/T}=T$ for some $\displaystyle k\in\mathbb Z^+.$ Then $\displaystyle g^kT=T$ and so $\displaystyle g^k\in T$ and so $\displaystyle (g^k)^m=g^{km}=e$ for some $\displaystyle k\in\mathbb Z^+.$ Hence $\displaystyle g$ is of finite order in $\displaystyle G,$ i.e. $\displaystyle g\in T$ so that $\displaystyle gT=T.$ This shows that $\displaystyle T$ is the only element of $\displaystyle G/T$ of finite order.