# Thread: Can you check my proof? (Torsion Subgroups of Abelians Groups are actual subgroups)

1. ## Can you check my proof? (Torsion Subgroups of Abelians Groups are actual subgroups)

I'm self studying Abstract Algebra Theory and Application and would appreciate a proof check, as I never invoke the fact that $\displaystyle G$ is an abelian group and the solution in the back of the book appears to be taking a different route than me, meaning my proof is likely wrong. If anyone could push me in the right direction I'd appreciate it.

Proposition: Let $\displaystyle G$ be an abelian group. The elements of finite order in $\displaystyle G$ form a subgroup $\displaystyle T$.

Proof. We know $\displaystyle T$ is not empty as the order of the identity of $\displaystyle G$ is one. We now must show that for all elements in $\displaystyle T$ there exists an inverse element. Consider any element $\displaystyle x$ of $\displaystyle T$ and let $\displaystyle m$ be its order. Thus, $\displaystyle x^m = e$, where $\displaystyle e$ is the identity of $\displaystyle G$ and now $\displaystyle T$. Hence, e=xm=x*xm-1=e, implying that xm-1 = x-1.
The binary operation on T is associative as it is already associative on G.

Quick notation clear up - * represents a binary operation.

Thanks!

2. ## Re: Can you check my proof? (Torsion Subgroups of Abelians Groups are actual subgroup

Originally Posted by Xyelos
I'm self studying Abstract Algebra Theory and Application and would appreciate a proof check, as I never invoke the fact that $\displaystyle G$ is an abelian group and the solution in the back of the book appears to be taking a different route than me, meaning my proof is likely wrong. If anyone could push me in the right direction I'd appreciate it.
Proposition: Let $\displaystyle G$ be an abelian group. The elements of finite order in $\displaystyle G$ form a subgroup $\displaystyle T$.
Proof. We know $\displaystyle T$ is not empty as the order of the identity of $\displaystyle G$ is one. We now must show that for all elements in $\displaystyle T$ there exists an inverse element. Consider any element $\displaystyle x$ of $\displaystyle T$ and let $\displaystyle m$ be its order. Thus, $\displaystyle x^m = e$, where $\displaystyle e$ is the identity of $\displaystyle G$ and now $\displaystyle T$. Hence, e=xm=x*xm-1=e, implying that xm-1 = x-1.
The binary operation on T is associative as it is already associative on G.Quick notation clear up - * represents a binary operation.
Here is a very useful characterization of subgroups: If $\mathcal{H}$ is a subset of a group $\mathcal{G}$ then $\mathcal{H}$ is a subgroup of $\mathcal{G}$ if and only if $\{a,b\}\subseteq \mathcal{H}$ then $ab^{-1}\in\mathcal{H}$.

In the posted question, you are asked to prove the the set of all elements of finite order form a subgroup of $\mathcal{G}$.
Do you know if the product of two elements of finite order has finite order?
Do you know if an element has finite order has finite order then its inverse has finite order?
Can these three facts be used to do your problem?