# proving closure of a subset [duplicate]

This question already has an answer here:

Let B be a set, and let * be a binary operation in B. Suppose * satisfies the associative law. Let

$$P=\{b \in B : b * w = w * b \quad\forall\, w \in B\}$$

Prove that P is closed under *.

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## marked as duplicate by Andrew Salmon, martini, Tomás, Gina, HaydenJul 21 '14 at 15:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

And as seen there, simply writing down what one has to show almost completes the proof. –  Hagen von Eitzen Jul 21 '14 at 13:52

## 1 Answer

Let $a,b\in P$, and we would like to show that $ab\in P$.

Since $a,b\in P$, we have that $a*w=w*a$ and $b*w=w*b$. Now, observe that \begin{align*} a*w&=w*a\\ a*w*b&=w*a*b \end{align*} Now, use $b*w=w*b$ on the first term.

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