The question could be trivial, but I want to know whether my arguments are correct.
Let $G$ be a finite group, $f$ be an automorphism of it. We say that $f$ is fixed-point free, if $f(x)=x$ implies $x=1$.
Let $N$ be $f$-invariant normal subgroup of $G$.
Question. If $f$ acts fixed-point freely on $N$ as well as $G/N$, can we conclude that it acts fixed-point freely on $G$?
My answer (not sure) yes. Because, $f$ is permuting elements of $N$, and the disjoint cosets of $N$. So, if $f(x)=x$, then $f(xN)=xN$; hence $xN=N$ (by fixed-point freeness on $G/N$). Thus $x\in N$. Then $x=1$ (by fixed-point freeness on $N$).
Is this argument correct?