# Proving a subset is closed under a binary operation on a set

Suppose that $*$ is an associative and commutative binary operation on a set $S$. Let $H=\{a\in S\mid a*a=a\}$. Show that $H$ is closed under $*$.

I started this problem by listing the definitions of * being commutative and associate. So I let $a,b\in H$. I now need to show that $a*b$ is in $H$. So I said $a*b=(a*a)*(b*b)$ because of what it means to be a member of $H$. I'm not sure where to go from there or how to prove it is associative. Any help?

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You want to prove that $(ab)(ab)=ab$. This is because $H$ consists of the objects $x$ that satisfy the equation $x^2=x$. You want to show that if $a$ and $b$ satisfy that equation, so does $ab$. First use associativity. –  André Nicolas Feb 7 '12 at 17:59
as you stated you should prove that ab in H if a and b in H. ab in H means that (ab)*(ab)=(a*b). can you prove this? –  miracle173 Feb 7 '12 at 18:02

You don't need to prove $*$ is associative; you already know it is.

The only thing you need to show is that if $a$ is in $H$ and $b$ is in $H$, then $a*b$ is in $H$.

When is an object $x$ in $H$? $x$ is in $H$ if and only if $x$ is in $S$, and $x*x=x$.

So in order to show that $a*b$ is in $H$, you need to show that:

• $a*b$ is in $S$; and
• $(a*b)*(a*b) = a*b$.

Why is $a*b\in S$? Because $a\in S$, $b\in S$, and $*$ is a binary operation in $S$. Remember that a binary operation is a function of type $*:S\times S \to S$, so $a*b$ maps to some element of $S$.

Why is $(a*b)*(a*b)=(a*b)$? Well, we know $a\in H$, so we know $a*a = a$; we know $b\in H$, so we know $b*b=b$. Now use the fact that $*$ is associative and commutative to show that $(a*b)*(a*b)=a*b$, and this will show that if $a\in H$ and $b\in H$, then $a*b\in H$. That is what you need to show in order to show that $H$ is closed under $*$.

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Okay, how does this look: (a∗b) = a∗(b∗b) = (a∗b)∗b = (b∗a)∗b = [b∗(a∗a)]∗b = b∗[(a∗a)∗b] = b∗[a∗(a∗b)] = (b∗a)∗(a∗b) = (a∗b)∗(a∗b) –  user23793 Feb 7 '12 at 18:12
@user23793: Fine, though it seems like it would be simpler to do $(a*b)*(a*b) = (a*b)*(b*a) = a*((b*b)*a) = a*(b*a) = a*(a*b) = (a*a)*b = a*b$ –  Arturo Magidin Feb 7 '12 at 18:16