Induced binary operator

1. Feb 6, 2010

bennyska

1. The problem statement, all variables and given/known data
suppose that * is an associative binary operation on a set S. Let H = {a elementof S | a*x = x*a forall x elementof S}. Show that H is closed under *.

2. Relevant equations

3. The attempt at a solution
i don't really know where to begin. I know i need to show forall a,b elementof H, a*b elementof H. i know * is associative, and H is the subset of all commutive elements of S. (the book says, "we think of H as consisting of all elements of S that commute with every element is S." same thing?) where do i go from here? what does * being associative have to do with commutive elements?

2. Feb 6, 2010

vela

Staff Emeritus
Let $a, b \in H$. To show closure, you want to prove that $a*b \in H$. In other words, you want to show that for all $x \in S$, $(a*b)*x = x*(a*b)$. Make sense?

3. Feb 6, 2010

bennyska

so could i say:
since * is associative, then for all a, b, x in S, (a*b)*x = a*(b*x), and since H is the set of all commutative elements, then (a*b)*x = a*(b*x) = x*(b*a) = x*(a*b), and H is closed under S. does that make sense, or did i just reiterate what you said? ( i guess i want to reiterate what you said with more detail.)

4. Feb 6, 2010

vela

Staff Emeritus
Yeah, essentially, but you need to pay more attention to the details. The statement only holds for a, b in H, not in S generally, but it needs to hold for all x in S. Also, just do one step at a time so it's clear that each step is justified. When you wrote a*(b*x)=x*(b*a), how do you know x and a can switch places like that? You are using neither commutativity nor associativity alone, so it's not immediately apparent to the reader that the step is correct.