If G is an abelian group, show that the set of elements of finite order is a subgroup of G.

Let G be an abelian group and H be the set of elements of finite order.
(1) nonempty
Now e ∈ H, since $a^n$ = e, by definition of order, and |e| = 1 ∈ H. Thus, H is nonempty.
(2) Closure
Let a, b ∈ H, where |a| = k and |b| = m for all k, m ∈ G. Then $a^k$ = e and $b^m$ = e. So $a^{km} = (a^k)^m = e$ and $b^{km} = (b^k)^m = e$. Then $(ab)^{km} = e$ and $ab$ ∈ H with a finite order at most $km$. Hence, H has closure.
(3) Inverse
Let a ∈ H with |a| = k for all k ∈ G. Then $(aa^{-1})^k = e$. So $a^k(a^-1)^k = e$. Since $a^k = e$, then $(a^{-1})^k = e$. Thus $|a^{-1}| = k$ and $a^{-1}$ ∈ H. Hence, H has an inverse.
Therefore, H is a subgroup of G.

I think I messed up on the inverse part of my proof.

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    $\begingroup$ Hi, I see you continue to title all your questions a permutation of "<field of mathematics> <something> proof help." Can you please stop doing that and pick more descriptive titles? $\endgroup$ – rschwieb Feb 9 '14 at 21:19

Writing $|a|=k$ and $|b|=m$ for all $k,m\in G$ makes no sense, since the order of an element is a positive integer. Also, you haven't proven $(ab)^{km}=e$ explicitly. You need to use the fact that $(ab)^n=a^nb^n$ in an abelian group.

The inverse part is straightforward; suppose $a^n=e$. Then $$ e=a^{-n}a^n=(a^{-1})^ne\implies (a^{-1})^n=e. $$ Hence $a^{-1}\in H$ as well.


Inverse: It is sufficient to show $ka=0$ iff $k(-a)=0$ (and thus $|a|=|-a|$ ) . But inversion is an invertible function so it is sufficient to show $ka=0$ implies $k(-a)=0$ . Assume $ka=0$. $ka=2ka+(k)(-a)$ by repeated application of the identity axiom. But this is $0=2*0+k(-a), k(-a)=0 $.


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