suppose that * is an associative binary operation on a set S. show that the set H= { a element S such that a*x=x*a for all x element S} is closed under *.

Relevance
• Anonymous

Let p and q be members of H. We seek to show that p*q is in H.

Let x be any member of S. Then (p*q)*x=p*(q*x) by associativity. Next, p*(q*x)=p*(x*q) because q is in H. Next, p*(x*q)=(p*x)*q because of associativity. Next, (p*x)*q=(x*p)*q since p is in H. Finally, (x*p)*q=x*(p*q) because of associativity. Putting all the equations together, we get (p*q)*x=x*(p*q). But this just shows that p*q is in H.

• Thefreevariable, that's wrong: not every binary associative operation on a set gives us a group. For example, the operation "standard multiplication" on the natural numbers N is an associative binary operation, and so is addition, and none yields N as a group (in both cases we get what is called a semigroup)

Regards

Tonio

• The proof above is great, I just want to add some terminology.

As my abstract algebra professor would say, "Let's not be coy." (S,*) is a group, so let's call it G. Then H is a special subgroup of G called the center of G. The center consists of the elements in G which commute with all other elements in G. What you have proven in this exercise is that the center of a group is closed.