Closure under * for subset H of commutative elements in set S

[bennyska]

Homework Statement

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 *.

Homework Equations

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?

 

[vela]

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?

 

[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.)

 

[vela]

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.