Is $H:=\{a\in G: a * s = s * a,\forall s\in S\}$ really a subgroup?

Question

Let $(G, *)$ be a group and $S\subset G$. How can I show that

$$H:=\{a\in G: a * s = s * a,\ \forall s\in S\}$$ is a subgroup of $G$?

It is easy to show $H\neq \phi$ and $H$ is closed under $*$. But, how about the inversion?

Thanks.

Answer 1

If $a \in H$, then we know $$ a \ast s = s \ast a $$ for each $s \in S$. Let us multiply by $a^{-1}$ on both the right and the left, to get $$ a^{-1} \ast a \ast s \ast a^{-1} = a^{-1} \ast s \ast a \ast a^{-1} \,, $$ which simplifies to $$ s \ast a^{-1} = a^{-1} \ast s \,. $$ Since this holds for all $s \in S$, we have shown that $a^{-1} \in H$.

Answer 2

Yes, $H$ is closed under inversion.

Suppose that there is an $a \in H$ such that $a^{-1}$ is not in $H$. Then $a^{-1}*s \not = s*a^{-1}$ for some $s \in S$. But then right-multiplying both sides by $a$ would give (1) $(a^{-1}*s)*a \not = (s*a^{-1})*a$.

However, on the one hand $(a^{-1}*s)*a = a^{-1}*(s*a) = a^{-1}*(a*s) = a^{-1}*a*s = s$. So this implies that (2) $(a^{-1}*s)*a = s$. On the other hand $(sa^{-1})*a = s*(a^{-1}*a)=s$. So this implies that (3) $(s*a^{-1})*a = s$ as well.

There is no way that (1), (2), and (3) can all hold.