Let G be an abelian group. Show $\{ x \in G | x^n=e$ for some $n\in \mathbb{Z} \}$ is a subgroup of G.
$\textbf{Proof:}$ Assume G is an abelian group. So G has an identity element, each element has a unique inverse, G has closure, G is associative, and if $a,b \in G$ then $ab=ba$. Let $$A=\{ x \in G | x^n=e \, \space\text{ for some } \space n\in \mathbb{Z} \}$$ We need to show that A is a subgroup of G. (i.e. A has an identity element, each element has a unique inverse, and G has closure.
We don't need to show that A has an identity element because the group description shows us that $x^n$ is the identity element since $x^n=e$. We also know because G has closure a subset of G, A, will also have closure.
So what is left to show is that A has an inverse. By the definition of an inverse,
\begin{equation*}
\begin{aligned}
x^nx^{-1} &=e=x^n \iff x^{-1}=1\\
x^{-1}x^n &=e=x^n \iff x^{-1}=1\\
\end{aligned}
\end{equation*}
Hence A has an inverse. So A is a subgroup of G.
Is this correct?
Best Answer
A hint towards showing closure:
Suppose $a\in A$; then you know some power of $a$ is the identity, let's say $a^j=e$. Similarly, if $b\in A$, then there's some power of $b$ that's the identity, say $b^k=e$.
Now, since any power of the identity is the identity (why?), then for any $m$ it's the case that (for instance) $(a^j)^m=a^{jm}=e$.
What's more, since $G$ is abelian then we know that $(ab)^n = a^nb^n$. Can you find some value of $n$ that you know will be guaranteed to make both terms on the right the identity? (And can you see why this shows closure?)