Showing that $S = \{ b \in G : a^{-1} b a = b \}$ is a subgroup of $G$.

Question

Let's say we have a group $G$ and $a$ is an element of $G$. I want to show that the set $S = \{ b \in G : a^{-1} b a = b \}$ is a group. The hint is to find inverses.

Answer 1

Assuming that $b\in S$ (i.e., that $a^{-1}ba=b$), you want to show that the inverse $b^{-1}$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^{-1}b^{-1}a=b^{-1}$: $$a^{-1}b^{-1}a = a^{-1}b^{-1}a =\cdot 1=a^{-1}b^{-1}a\cdot \color{red}b\cdot b^{-1} = \underbrace{a^{-1}b^{-1}a\cdot \color{red}{a^{-1}ba}}_{\text{all cancels}}\cdot b^{-1}=b^{-1}$$