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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 $G$?

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It sounds like you mean for $S$ to be a subset. Also, elements such that $a^2=a$ are called idempotent, but elements such that $a^2=e$ are not. If you don't really need to define that (which it looks like you don't) you should probably leave it out. – rschwieb Dec 5 '12 at 18:14
Do you want $S$ to be the subset of $G$ that consists of the idempotent elements of $G$? – Chris Leary Dec 5 '12 at 18:14
@rschwieb: Wish you edit the title as well. :) – Babak S. Dec 5 '12 at 18:17
@BabakSorouh I would spend the time but I think this is probably a duplicate anyhow... – rschwieb Dec 5 '12 at 18:18
@rschwieb: Have a look at this link. when $H=\{e\}$. – Babak S. Dec 5 '12 at 18:20

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 $S$?

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I like your sake for learning the basic concepts here(As you noted in teaching complex numbers as well). :) – Babak S. Dec 5 '12 at 18:23
@julien I'm not clear what statements you're referring to. – amWhy Nov 6 '13 at 14:49
Wait, I had not seen the dates of the post and answers... Disregard my comments. Apologies. – 1015 Nov 6 '13 at 14:52
@julien My energy has since been sapped by such posts over the ensuing year (following this post), and the energy-sapping-ness isn't all-pervasive. What I find most energy-sapping are repeated users repeatedly posting PSQ's. – amWhy Nov 6 '13 at 14:53
@julien Yes, time wears on us all! No problem...hence I'm a little more cynical now than I was then! ;-) – amWhy Nov 6 '13 at 14:54

There are different ways of proving this statement. One possibility is to write $S = \ker f$ for a proper chosen homomorphism $f \colon G \to G$.

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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.

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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:

  1. $H$ is closed under $\ast$ as we verified above.
  2. $e\in H$ 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 $e$).
  3. If $a\in H$ then $a^{-1}\in H$.

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

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