Here is the problem:
Suppose that $*$ is an associative binary operation on a set $S$. Let
$$H:= \{a \in S\mid a * x = x * a \mbox{ for all }x\in S\}.$$
In other words, $H$ is consisting of all the elements of $S$ that commute with every element in $S$.
Show that $H$ is closed under $*$.
On
Alright, I think I got it. Using the start that @Michael Joyce gave me...
(1) Since ∗ is an associative operation, (a∗b)∗x=a∗(b∗x).
(2) Since b∈H, b∗x=x∗b, so a∗(b∗x)=a∗(x∗b).
(3) Since ∗ is an associative operation, a∗(x∗b)=(a*x)*b.
(4) Since a∈H, a∗x=x∗a, so (a*x)*b=(x*a)*b.
(5) Since ∗ is an associative operation, (x*a)b=x(a*b).
Since (a∗b)∗x=x*(a*b) then a*b ∈ H.
So, H is closed under *.
Suppose that $a,b \in H$. You must show that $a * b \in H$. Thus, you must show that for any $x \in S$, $(a * b) * x = x * (a * b)$.
Here are some hints to get you started:
(1) Since $*$ is an associative operation, $(a * b) * x = a * (b * x)$.
(2) Since $b \in H$, $b * x = x * b$, so $a * (b * x) = a * (x * b)$.
Can you finish the proof from here? You're going to use a series of steps that are each similar to either (1) or (2).