Let G is a group, $\phi$ is an automorphism, and $ A=\{ g\in G\mid \phi(g)=g^{-1} \}$. Consider $a \in A$ and $B_a =\{ g\in A\mid ga \in A \}$.
Prove that $g \in B_a$ implies $g^{-1} \in B_a$.
It seems this problem can be solved directly. However, I could not solve it without using centralizer subgroups.
This is my solution:
Lemma: $A \cap Z(a) =B_a$, where $Z(a)$ is the centralizer of $a$.
Proof: $g \in B_a \Rightarrow ga \in A \Rightarrow \phi(ga)=(ag)^{-1}$. Also $\phi(ga)=\phi(g) \phi(a)=g^{-1} a^{-1}=(ga)^{-1}$. So, $ag=ga$ and $B_a \subset Z(a)$.
Since $B_a \subset Z(a)$ and $B_a \subset A$, then $A \cap Z(a) =B_a$. $\blacksquare$
$g \in A$ implies $g^{-1} \in A$, and $g \in Z(a)$ implies $g^{-1} \in Z(a)$. Therefore, with using the lemma, $g \in A \cap Z(a)=B_a$ implies $g^{-1} \in B_a$.
Now, I am wondering why I cannot solve it using the following way:
$g \in B_a$ implies $ga=ag \in A$. Also, $ga^{-1}=a^{-1}g$. Now, in order to prove $g^{-1} \in B_a$, I only need to prove $g^{-1}a \in A$ or $ga^{-1} \in A$, but I cannot.
Does anyone have another solution or know how to continue this approach?
You can essentially do it the second way, withan adjustment: you have that $g\in B_a\implies ga=ag\implies a^{-1}g=ga^{-1}$. That's not too hard to get.
Now, we need to prove $g^{-1}a\in A$. So look at $\phi(g^{-1}a)=\phi(g^{-1})\phi(a)=ga^{-1}=a^{-1}g=(g^{-1}a)^{-1}$.
Thus $g^{-1}a\in A$. So $g^{-1}\in B_a$.