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 $G$?
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 $G$?
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 $e \in S$?
$(2)$ For each $a \in S$, is $a^{-1} \in S$?
$(3)$ For each $a, b \in S$, is $a * b \in S$?
Since the answer to all three questions is "yes", then what can we conclude about $S$?
There are different ways of proving this statement. One possibility is to write $S = \ker f$ for a proper chosen homomorphism $f \colon G \to G$.
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.
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:
Therefore $H$ is a subgroup of $G$.