# Is associative binary operator closed on this subset?

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 $*$.

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What have you tried? What do you think you need to show? –  Thomas Andrews Sep 17 '13 at 15:43
I thought that by the definition of a binary operation * on S, the set S was closed under *. Doesn't that mean the subset H is also closed? –  Jesus Sep 17 '13 at 16:02
No, it is "closed" on $H$ if for each $a,b\in H$, $a*b\in H.$ All you know from knowing $*$ is closed on $S$ is that $a*b\in S.$ –  Thomas Andrews Sep 17 '13 at 16:49

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).

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So, I would have to show that every combination of a, b, x satisfies the associative operation? –  Jesus Sep 17 '13 at 15:50
You know that the associative property holds for all elements of $S$. What you need to show is that if you assume that $a$ and $b$ belong to $H$, then the element $a * b$ (which a priori is only known to be an element of $S$) is also in $H$. –  Michael Joyce Sep 17 '13 at 16:04
So what else is left to prove if you showed that the associative operation holds for a and b? –  Jesus Sep 17 '13 at 16:06
You must prove "If $a, b \in H$, then $a * b \in H$." This is NOT immediate and requires you to use both the associativity property of the operation (1) and the definition of the set $H$ (2). –  Michael Joyce Sep 17 '13 at 16:28

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 *.

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Ah, your right. I guess changed it. –  Jesus Sep 19 '13 at 11:49