Proving abelian subgroups: $S = \{a \in G\mid a^2 = e\}$ Let $(G,*)$ be an abelian group with the identity $e$. An element $a\in G$ is called an idempotent if $\,a^2 = e\,$ (where $\,a^2 = a*a).\,$ Let $S = \{a \in G\mid a^2 = e\}.$ 
How do I prove $S$ is a subgroup of $G$?
 A: 
Let $(G,*)$ be an abelian group with the identity $e$. Let $S$ be the set of all elements $a\,\in G\,$ where $\,a^2 = a*a =e$. We need to prove that $S\le G$.

$(1)$ Is $e \in S$?


*

*We have $e$ as the identity in $G$. Now, $e^2 = e * e = e$.So $e \in S.$So $S$ is non-empty.


$(2)$ For each $a \in S$, is $a^{-1} \in S$? 


*

*Let $a \in S$ be any element in $S$. So we know $a^2 = e.\;\;$
Then $(a^{-1})^2 = (a^{-1})*(a^{-1}) = (a^{-1})*e*(a^{-1}) = (a^{-1})*a^2*a^{-1}$  $\quad\quad\quad = a^{-1}*(a*a)*a^{-1} = (a^{-1}*a)*(a*a^{-1}) = e * e = e.$ So $a^{-1} \in S$.


$(3)$ For each $a, b \in S$, is $a * b \in S$?


*

*Let $a \in G$ and $b \in G$, so that $a^2 = a*a=e, \;\; b^2 = b*b = e$.

Then $(a*b)^2 = (a*b)*(a*b) = (b*a)(a*b) = b*(a^2)*b = b*e*b = b*b =e$. 
$\quad\quad\quad\quad $ (Recall, $G$ is abelian. $a, b \in S \implies a, b \in G \implies a*b = b*a$). 
So $(a*b)\in S$.



Since the answer to all three questions is "yes", then what can we conclude about $S$?
A: There are different ways of proving this statement. One possibility is to write $S = \ker f$ for a proper chosen homomorphism $f \colon G \to G$. 
A: One way would be to let $a$ and $b$ be idempotent and show that $a*b^{-1}$ is also idempotent. Definitions and the fact that $G$ is abelian should get you through from here.
A: How do you prove anything is anything? You verify that a list of definitions hold, either directly or by "alternative characterizations".
For example since $G$ is abelian we know that every subgroup is normal, so it is enough to show that this collection is the kernel of a homomorphism. You could verify that the product of idempotents is idempotent, and so is the inverse. And so on. Pick your pick.
Note that if $a$ is an idempotent then $a^{-1}=a$, so it is enough to show that the product of idempotents is idempotent. Suppose that $a$ and $b$ are idempotent we wish to show that $(a\ast b)^2=e$. We calculate: 
$$(a\ast b)^2=a\ast b\ast a\ast b=a\ast a\ast b\ast b=a^2\ast b^2=e\ast e=e$$
So the product of idempotent is idempotent, and by definition idempotents are their own inverses, so we have that if $H$ is the set of all idempotents then:


*

*$H$ is closed under $\ast$ as we verified above.

*$e\in H$ because $e\ast e=e$ (note that it is tempting to deduce that $a\ast a=e$ but it might be the case where $e$ is the only idempotent, in which case we will assume the result we wish to show: the idempotency of $e$).

*If $a\in H$ then $a^{-1}\in H$.


Therefore $H$ is a subgroup of $G$.
