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Let $G$ be an Abelian group. And let H={$g\in G |\mathrm{order}(g) < \infty$}

I need to prove that H is a normal subgroup in G.

I know that if i prove it's a subgroup, it will be normal as well, since G is Abelian. But i wonder about it being a subgroup.

I know this has to work: $$\forall h_1, h_2\in H,\hspace{1cm}h_1*h_2^{-1}\in H$$

But I wonder, is it enough to just show that the order of the product is still finite? Is there any special way to show the order is finite? Or is it just trivial?

If anyone can help clearing this out, I would really appreciate it.

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    $\begingroup$ I wouldn't say this is trivial, since it is false for non-abelian groups. You need to use the fact that since $G$ is abelian, $(ab)^n=a^nb^n$ for any $a,b\in G$. . $\endgroup$ – Mathmo123 Feb 9 '17 at 11:22
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If $a^n=1$ then $(a^{-1})^n$ is also $1$, because $(a^{-1})^na^n=1$ (the last can be proved by induction on $n$). If $a^n=1$ and $b^m=1$ then $(ab)^{nm}=a^{nm}b^{nm}=(a^n)^m(b^m)^n=1$. (Note that commutativity is used in the first equality.) Hence if $a$ and $b$ are of the finite order then $a^{-1}$ and $ab$ are also of the finite order.

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As you remarked correctly it is sufficient to prove $h_1^{-1}h_2 \in H$. Let $\lvert h_1 \rvert = p_1$ and $\lvert h_2 \rvert = p_2$. If $p = p_1p_2$ then $h_1^p = h_2^p = (h_1^{-1})^p = 1$ and so also $(h_1^{-1}h_2)^p = 1$.

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