# group element order

• May 1st 2009, 05:19 AM
mpryal
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!
• May 1st 2009, 05:50 AM
hatsoff
Quote:

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 $a,b\in T$. Then there exists positive integers $m,n$ with $a^m=e$ and $b^n=e$, with $m,n>1$. Since $a,b\in G$, then $ab\in G$. But is it also in $T$? Well, observe that $(ab)^{mn}\in G$. Also, $(ab)^{mn}=a^{mn}b^{mn}=(a^m)^n(b^n)^m=e^ne^m=e$. So then the conditions are satisfied such that $ab\in T$, and therefore $T$ is closed.

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

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

That should do it.
• May 1st 2009, 07:23 AM
TheAbstractionist
Quote:

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