Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
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 at 8:15
    
Thanks, I really feel stupid for not seeing this simple example... –  OliverX1 Apr 17 at 7:25

1 Answer 1

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 $N$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.