Proving Finite Order Elements Form a Subgroup of an Abelian Group

[rideabike]

Homework Statement

Prove the collection of all finite order elements in an abelian group, G, is a subgroup of G.

The Attempt at a Solution

Let H={x\inG : x is finite} with a,b \inH.
Then a^{n}=e and b^{m}=e for some n,m.
And b^{-1}\inH. (Can I just say this?)
Hence (ab^{-1})^{mn}=a^{mn}b^{-mn}=e^{m}e^{n}=e (Since G is abelian the powers can be distributed like that)

So ab^{-1}\inH, and H≤G.

 

[jbunniii]

Homework Statement

Prove the collection of all finite order elements in an abelian group, G, is a subgroup of G.

The Attempt at a Solution

Let H={x\inG : x is finite} with a,b \inH.
Then a^{n}=e and b^{m}=e for some n,m.
And b^{-1}\inH. (Can I just say this?)

If you're going to say it, you should justify it. But your proof below doesn't use this fact. (Indeed, it proves it!)

Hence (ab^{-1})^{mn}=a^{mn}b^{-mn}=e^{m}e^{n}=e (Since G is abelian the powers can be distributed like that)

So ab^{-1}\inH, and H≤G.

This part is fine. Note that as a special case, this shows that $b^{-1} \in H$. (Take $a = e$.) You didn't need to stipulate $b^{-1}\in H$ in the previously quoted section.

 

[Evo]

Delete thread, figured it out

We don't delete threads once they have a response.

Thank you jbunniii!