# Induced automorphism on quotient group

I am stuck on a problem studying for my prelims : Let $G$ be a group, $N \unlhd G$ a normal subgroup and $\alpha \colon G \to G$ an automorphism of $G$ such that $\alpha(N) \leq N$. The first part is to prove that $\alpha$ induces an automorphism $\overline{\alpha}$ of $G / N$. The second part is to find an example where $\alpha$ restricted to $N$ and $\overline{\alpha}$ are both identities but $\alpha$ is not the identity.

I am stuck in the first part... So since $\alpha (N) \subseteq N$, there is an induced endomorphism $\overline{\alpha} \colon G/N \to G/N$, which clearly is surjective. However, I cannot prove the injectivity, which is equivalent to proving that there is an induced hm $\overline{\alpha^{-1}}$, which is the same as $\alpha^{-1} (N) \leq N$.

We would be done if the group $G$ is assume to be finite for example, but I don't see how to prove it general. Thanks for your help

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This is not true in general. Consider $G=\mathbb{Q}$, $N=\mathbb{Z}$, and $\alpha:\mathbb{Q} \to \mathbb{Q}:x\mapsto2x$. Then $\alpha(N) = 2\mathbb{Z} \leq N$, and $\overline{\alpha}:\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}:x\mapsto 2x$ remains surjective, but is not injective.

The question should have hypothesized that $\alpha(N) = N$, or as you observed, have additionally hypothesized $\alpha^{-1}(N) \leq N$ (which amounts to the same thing).

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Thanks, that is what I was secretly hoping for, but didn't find a couterexample. I don't know why I was looking for a non-abelian example! Thanks a lot :) – Bogdan Sep 29 '12 at 20:27
Jack Schmidt, there's a problem in Herstein's Topics in Algebra, 2nd edition, which also asks the reader to prove that $\sigma$ induces an automorphism of the quotient group $G/N$ whenever $N$ is a normal subgroup of $G$ such that $\sigma(N) \subset N$, where $\sigma$ is an automorphism of $G$. . It's Problem 19 in Section 2.8. – Saaqib Mahmuud Jun 16 '14 at 14:51

Since the group $\mathbb{Z}/2$ has only trivial automorphism, we would like to construct group $G$ with a (normal) subgroup and quotient (both) isomorphic to $\mathbb{Z}/2$.

The smallest example is $G=\langle x\colon x^4\rangle\cong \mathbb{Z}/4$, and let $\alpha\colon x\mapsto x^3$.

The smallest non-abelian group is $S_3$. It has presentation $\langle x,y\colon x^3,y^2,y^{-1}xy=x^{-1}\rangle$. The only normal subgroup is $\langle x\rangle$, and (interestingly) quotient is $\mathbb{Z}/2$, set $\alpha(x)=x$ (as you want). We shouldn't have $\alpha(y)\neq y$ (for $\alpha\neq I$). Set $\alpha(y)=xy$.

Check that $\alpha$ is a required automorphism.

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Thanks for the examples :) – Bogdan Oct 3 '12 at 2:12