Let $G$ be a group, with $N$ characteristic in $G$. As $N$ is characteristic, every automorphism of $G$ induces an automorphism of $G/N$. Thus, $\operatorname{Aut}(G)\rightarrow \operatorname{Aut}(G/N)$. I was therefore wondering,

Under what conditions is the induced homomorphism $\operatorname{Aut}(G)\rightarrow \operatorname{Aut}(G/N)$

  • a monomorphism?

  • an epimorphism?

  • an isomorphism?

I believe it should work for (semi-?)direct products $N\times H$ where $\operatorname{Aut}(N)$ is trivial and $N\not\cong H$ (for example, $C_2\times C_3$, $N=C_2$). But I can't prove even that!

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    $\begingroup$ @Michael Hardy: (I ask this question out of interest, and is not meant to be hostile in any way, shape or form!) Why did you edit my question to add \operatorname before each operator? My understanding of \operatorname in Latex is that it is purely asthetic...was this the reason you changed it, or is there something more complex going on? (I would have sent you a private message to ask this, but I cannot seem to work out how too...) $\endgroup$
    – user1729
    Sep 14, 2011 at 14:09
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    $\begingroup$ What's wrong with aesthetics? Writing the operators in the default font (which is kerned to make juxtaposed letters look like they're variables being multiplied) rather than an upright font with word kerning is akin to a spelling error. Usually, editing not-own questions to correct trivial typos is frowned upon, but an exception is generally recognized for TeXnicalities, because it's thought to be helpful to the asker (and future readers) to show how things ought to be formatted. $\endgroup$ Sep 14, 2011 at 15:07
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    $\begingroup$ Nothing is wrong with aesthetics. As I said, I was just wondering if there was something more complex going on. $\endgroup$
    – user1729
    Sep 14, 2011 at 15:09
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    $\begingroup$ To add to @Henning 's point: there is also a semantic aspect to it. While these purely aesthetic things may seem like quibbles, there is the point that the engines like css, markdown, mathjax, and whatever else is used on this site have no clue what they are displaying, they're just trying to make some sense out of what you feed them. By using proper formatting you ensure that whatever design tweaks are incorporated in the future, the results should still remain human readable. So, for example, your list would be better displayed if you added a blank after the minus signs. $\endgroup$
    – t.b.
    Sep 14, 2011 at 15:32
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    $\begingroup$ @jug: That obvious map has an interesting kernel (1-cocycles from G to H=G/N). A very well written and interesting paper is that of Curran: 'Automorphisms of Semidirect Products' (link: jstor.org/discover/10.2307/…) There you actually get a bit more of a generalization: if the subgroup N is characteristic, Curran's construction gives you the full automorphism group of the product G=HN, but if it is just normal and G fixes it set-wise, his construction also gives you the full automorphism groups. $\endgroup$
    – user34168
    Jun 21, 2012 at 13:28

1 Answer 1


For direct product it seems easy.

Let $G=N\times K$ and assume that both $N$ and $K$ are characteristic in $G$. It is easy to show $\text{Aut}(G) \cong \text{Aut}(N)\times \text{Aut}(K)$ since both $N$ and $K$ are characteristic in G. Since $G/N \cong K$ then $\text{Aut}(G/N) \cong \text{Aut}(K)$. Thus there is a natural epimorphism $\phi:\text{Aut}(G) \to \text{Aut}(G/N)$ with $\text{Ker}(\phi) \cong \text{Aut}(K)$.

Now you can ask when are they both characteristic in $G$? Actually, one simple condition provide this: Let $N$ and $K$ be finite groups with relatively prime orders, and set $G=N\times K$. Then both $N$ and $K$ are characteristic in $G$.

And you offer an example $G=C_2\times C_3$ and $N=C_2$ then set $K=C_3$ since order of N and K are relatively prime, the result is immediate from above construction.


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