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?
Best Answer
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:
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 $*$.