Show for some subsets of $G$ we have subgroups of $(G, \ast)$ Let $G$ be an abelian group. Show that for the following subsets $H_n$, we have subgroups of $G$.
$H_1= \lbrace g \in G | g^n=e \rbrace $, with $n$ being a certain fixed natural number.
$H_2 = \lbrace g \in G | g^{-1}=g \rbrace$
$H_3 = \lbrace g \in G | g=x^2 $ for a $x \in G \rbrace$

For $H_1$:
$e \in H_1$ is obvious.
Let be $k \in H_1$, so $k^n=e\Longleftrightarrow k \ast(k)^{n-1}=e\Longleftrightarrow k^{-1}=(k)^{n-1}$
We need to show that $(k)^{n-1} \in H_1$
So we show:
$((k)^{n-1})^n=e$
$((k)^{n-1})^n=\underbrace{k^{n-1}\ast k^{n-1} \ast...\ast k^{n-1}}_{n}=\overbrace{\underbrace{k^{n}\ast k^{n} \ast...\ast k^{n}}_{n-2}\ast \underbrace{k^{n-1}\ast k^{1}}_{=e}}^{\text{Since $\ast$ 
is associative}}=\underbrace{e\ast e \ast...\ast e}_{n-2}\ast e=e$
$\Longrightarrow \forall k \in H_1:k^{-1} \in H_1$
We show $\forall k,t \in H_1: k\ast t \in H_1$:
To show that we need to show: $(k \ast t)^n = e$
$(k \ast t)^n=\underbrace{(k \ast t) \ast (k \ast t) \ast ... \ast (k \ast t)}_{n}=\overbrace{\underbrace{(k \ast... \ast k \ast k)}_{n} \ast \underbrace{(t \ast ...\ast t \ast t)}_{n}}^{\text{since $(G,\ast)$ is associative and kommutative}}=k^n\ast t^n=e \ast e= e$
$\Longrightarrow \forall k,t \in H_1: k\ast t \in H_1$
$\Box$
For $H_2$:
$e \in H_2$ is obvious.
Let $k \in H_2 \Longrightarrow k=k^{-1}\Longrightarrow \forall k \in H_2:k^{-1} \in H_2$
We now show that $\forall k,t \in H_2: k \ast t \in H_2$:
In order for $k \ast t \in H_2$, $\,\,\,(k \ast t)=(k \ast t)^{-1}$ has to hold!
Here again the kommutativity of $(G,\ast)$ plays a role!
$k \ast t \ast k^{-1} \ast t^{-1}= k \ast t \ast t^{-1} \ast k^{-1}=k \ast e \ast k^{-1}= k \ast k^{-1}=e$
This tells us indeed: $(k\ast t)^{-1}=k^{-1}\ast t^{-1}=k \ast t= (k \ast t)$
$\Box$
For $H_3$:
Again $e=e^2 \Longrightarrow e \in H_3$
Let $k \in H_3 \Longrightarrow k = x^2$ for some $x \in G$
Then $k^{-1}=(x^2)^{-1}$ which is again since we have an abelian group $(x^2)^{-1}=(x^{-1})^2=k^{-1}$
$\Longrightarrow \forall k \in H_3:k^{-1} \in H_3$
We now show $\forall k,t \in H_3: k\ast t \in H_3$:
$k \ast t= x^2 \ast y^2$ with $x,y \in G$
$k \ast t= x^2 \ast y^2=x \ast x \ast y \ast y= x \ast y \ast x \ast y=(x\ast y)^2 \longleftarrow$ because its still an abelian group
Since $x,y \in G \Longrightarrow x \ast y \in G$
Let $(x \ast y):= z$
$\Longrightarrow k \ast t=z^2$
$\Longrightarrow \forall k,t \in H_3: k\ast t \in H_3$
$\Box$

It would be great if someone could look over it and give me some feedback :)
 A: For $H_1$ note that
$$g^n=e\implies \left(g^n\right)^{-1}g^n=\left(g^n\right)^{-1}\implies \left(g^{-1}g\right)^n=e=(g^{-1})^n$$
That $\left(g^{-1}\right)^n=\left(g^n\right)^{-1}$ can be deduced from inverses being unique (this implies that the order of an element and of its inverse coincide). Your proof is fine too but a bit lengthy.
Besides that I cannot spot any significant shortcuts based on a case-by-case consideration. Anyway, the following known as One-Step subgroup test (EDIT: I realized, you are aware of this) might be of interest

Claim. A non-empty subset $H$ of a group $G$ is a subgroup if and only if $a,b\in H\implies a\circ b^{-1}\in H$.

Proof. $H$ being subgroup implying the latter condition should be clear. For the converse take $x\in H$ (there is such an $x$ as $H$ is non-empty) and let $a=b=x$. Then $a\circ b^{-1}=x\circ x^{-1}=e\in H$. Now, take $a=e$ and $b=x$ to deduce that for all $x\in H$ we have $a\circ b^{-1}=e\circ x^{-1}=x^{-1}\in H$. Finally, for $x,y\in H$ we have $y^{-1}\in H$ and thus $a=x$, $b=y^{-1}$ implying $a\circ b^{-1}=x\circ(y^{-1})^{-1}=x\circ y\in H$ finishing the proof.$~~~\square$
This is a useful criterion which usually shortens up the amount of calculations necessary. Take $H_1$ and oberserve that for $g,h\in H_1$ we have $g^n=h^n=e$ and thus
$$(g\circ h^{-1})^n=(g\circ h^{-1})\circ\dots\circ(g\circ h^{-1})=g^n\left(h^{-1}\right)^n=e$$
And as $e\in H_1$ the One-Step subgroup test yields the result (arguably, more calculations are hidden within the fact $\left(h^{-1}\right)^n=\left(h^n\right)^{-1}$ if not yet established). I encourage you to try it for $H_2,H_3$ too!

Minor spelling remark: it is written 'commutative' in English, not 'kommutative'. Anyway, as a German native speaker I understand the tendency towards writing the latter.
