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Assume we have a group extension $1 \to N \to G \to H \to 1$, and an automorphism $\phi: G \to G$. Is it correct that this automorphism induces automorphisms $\phi_N : N \to N$ and $\phi_H : H \to H$ ?

If so, this would mean that the image by $\phi$ of elements of the form $(n,1_H) \in G$ are the elements $(\phi_N(n),1_H)$, and that elements of the form $(n,h) \in G$ are sent to $(\phi_N(n)\cdot n'(h),\phi_H(h))$, where $n'(h)$ is an element of $N$ which depends on $h$. Is this also correct ?

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There could be automorphisms of $G$ that do not fix the image of the subgroup $N$. For example, if $G = C_2 \times C_2$ with $|N|=|H|=2$, then $G$ has an automorphism of order $3$. – Derek Holt Apr 16 '14 at 8:15
Thanks, I really feel stupid for not seeing this simple example... – OliverX1 Apr 17 '14 at 7:25
up vote 1 down vote accepted

Not true. Take an automorphism which is not inner. For abelian examples, take $G$ to be the points of plane under vector addition, and $N$ to be any line through origin.

Now rotations of the plane by any angle (not $0$ or $\pi$) is an automorphism which does not take $N$ to $N$.

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