Showing $f(h)=g^{-1}h^{-1}gh$ for any $g \in G, h\in H$ where $H$ is normal subgroup of $G$.

Question

Let $H$ be a finite normal subgroup of $G$. Show that for any $g\in G$, the map $f:H \to H$ defined by $f(h)=g^{-1}h^{-1}gh$ is bijective.

Edit to problem:

Let $H$ be a finite normal subgroup of $G$. Let $g\in G$ have order $n$ and the only element in $H$ that commute with $g$ is $e$, the identity element of $H$.

Show that for any $g\in G$, the map $f:H \to H$ defined by $f(h)=g^{-1}h^{-1}gh$ is bijective.

Here's my attempt. I can't prove the surjectivity yet. So, I try prove the injectivity.

Let $a,b \in H$. Assume that $f(a)=f(b)$. Now,

$g^{-1}a^{-1}ga = g^{-1}b^{-1}gb$

$a^{-1}ga = b^{-1}gb$

$a^{-1}gag^{-1} = b^{-1}gbg^{-1}$

$a^{-1}H = b^{-1}H$

$a^{-1} = b^{-1} \Leftrightarrow a=b$

Is that true? If no, how to prove it? Any idea to prove the surjectivity too? Thanks in advance.

Answer 1

You cannot conclude from $a^{-1}H = b^{-1}H$ that $a^{-1}=b^{-1}$. For one thing, since you are assuming that $a,b\in H$, then you always have $a^{-1}H=b^{-1}H = H$, whether $a$ and $b$ are equal or not.

However, you are almost there in the modified problem, with $H$ finite and the assumption that no element of $H$ other than $e$ commutes with $g$. From $f(a)=g(b)$, you have $a^{-1}ga = b^{-1}gb$. This gives you $$\begin{align*} a^{-1}ga &= b^{-1}gb\\ ba^{-1}g &= gba^{-1}\\ (ba^{-1})g &= g(ba^{-1}). \end{align*}$$ Can you take it from there?

Answer 2

This was the answer to the original question. The statement is wrong. For example, if $H$ is a central subgroup (e.g., the center of $SL_2(\mathbb{Z})$) then the map is a constant.