Let $f:G \rightarrow G$ is automorphism. If $N(g) = \lbrace x \in G | xg=gx, \forall g \in G \rbrace$. Prove that $f(N(g)) = N(f(g))$.

Question

Let $G$ be a group and $f:G \rightarrow G$ is automorphism. Define a set $N(g) = \lbrace x \in G | xg=gx, \forall g \in G \rbrace$.

Prove that $f(N(g)) = N(f(g))$.

Here I tried.

Let's define $f:G \rightarrow G$ by$ f(x)=g^{-1}xg$. We'll show that $f(N(g))=N(f(g))$.

First, we'll show that $f(N(g)) \subseteq N(f(g))$. Note that $xgx^{-1}=g \in N(g)$. Consider that $f(N(g)) = g^{-1}(N(g))g = g^{-1}(xgx^{-1})g = g^{-1}(gxx^{-1})g = g = xgx^{-1} (definition) = x(g^{-1}gg)x^{-1} = x(f(g))x^{-1} \in N(f(g))$

Conversely, $N(f(g)) = x(f(g))x^{-1} = x(g^{-1}gg)x^{-1} = xgx^{-1} = g = g^{-1}gxx^{-1}g = g^{-1}xgx^{-1}g = g^{-1}(N(g))g \in f(N(g))$

Thus, $f(N(g)) \subseteq N(f(g))$ and $N(f(g)) \subseteq f(N(g))$. Hence, $f(N(g)) = N(f(g))$.

Thanks for help and correction in advanced.

Answer 1

You can't just pick an automorphism and show that it works. You need to prove the result for an arbitrary automorphism.

Suppose $x\in N(g)$, so that $xg=gx$. Then $f(xg)=f(x)f(g)=f(g)f(x)=f(gx)$. Thus $f(x)\in N(f(g))$, so $f(N(g))\subseteq N(f(g))$.

Suppose $y\in N(f(g))$, so that $yf(g)=f(g)y$. Then $f^{-1}(yf(g))=f^{-1}(y)g = gf^{-1}(y)=f^{-1}(f(g)y)$, so $f^{-1}(y)\in N(g)$. Thus $y\in f(N(g))$ since $f(f^{-1}(y))=y\in N(f(g))$ and $f^{-1}(y)\in N(g)$, so $N(f(g))\subseteq f(N(g))$.