Check: Let G be an abelian group. Show $\{ x \in G | x^n=e$ for some $n\in \mathbb{Z} \}$ is a subgroup of G. 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?
 A: 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?)
A: A subset of a group doesn't necessarily have closure. To show closure, you need to select two arbitrary elements of $A$, say $x, y \in A$, and show that $xy \in A$.
Let $x, y \in A$. Then $x^n = e, y^n \in e$, where $n\in \mathbb Z$. Then $x^ny^n = ee = e = (xy)^n$, because $G$ is abelian. Hence, $xy\in A$. Thus, $A$ is closed under the group operation of $G$ (and hence under the group operation of $A$).
To show that $A$ is closed under inverses, you need to show that for an arbitrary element $x\in A$, that $x^{-1} \in A$ as well. (We know that there exists an inverse $x^{-1} \in G$, we just need to show that it is also in $A$ if $A$ is a subgroup of $G$.
Let $x \in A$. So, $x^n = e$. Hence $(x^{-1})^n = (x^n)^{-1} = e^{-1} = e$. Indeed, $x^{-1}\in A$.
A: We need to show three properties of a subgroup:
(a) Identity element e $\in$ A
(b) There exists $h^{-1} \in A$ for every $h \in A$
(c) For every $g, h \in A$ the product $gh \in A$
For (a), we know that $g^n \in A$ and $g^n = e$ so e $\in$ A. For (b), we can show that $h^{-1} = (g^n)^{-1} = e \in A$ and thus $h^{-1} \in A$.
Finally, for (c) we can show that if $g^n \in A$ and $h^n \in A$, and the group G is abelian, then it satisfies the commutative property $g^nh^n = (gh)^n = e \in A$. So A is closed under the product operation. Thus we have satisfied all three properties and A is a subgroup of G.
