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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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    @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...)2011-09-14
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    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.2011-09-14
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    Nothing is wrong with aesthetics. As I said, I was just wondering if there was something more complex going on.2011-09-14
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    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.2011-09-14
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    Your condition on $N$ and $H$ in the final comment are not enough even to guarantee that $N$ is characteristic, so that may be why you are having trouble. For example, you can have $N=C_2$, $H=C_2\times C_2$, and $G=N\times H$ (viewed as a semidirect product). Clearly, $N\not\cong H$, but $N$ is not characteristic in $G$.2011-09-14
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    @Arturo Magidin: Sorry - I was assuming $N$ to be characteristic.2011-09-15
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    @Swlabr: (about your case of a semi-direct product) Do you know any groups except $1$ and $Z_2$ with trivial automorphism group?2011-09-15
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    No! Which is a good point. Such a group is necessarily abelian (as conjugation is always an automorphism), but f.g. abelian groups will always have automorphism groups as you can just use their categorisation and send $x\mapsto x^{-1}$ in one of the cyclic subgroups in the decomposition, or switch two $C_2$s. So yes, there are only two groups with trivial automorphism group...hmm...2011-09-15
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    @Swlabr: If $N$ contains a non-central element $n \in N\setminus \operatorname{Z}(G)$, then the non-trivial inner automorphism induced by $n$ maps to the trivial automorphism on $G/N$. Hence your map is not one-to-one. Maybe it's more natural to look at the map $\operatorname{Aut}(G) \to \operatorname{Aut}(G/N)\times\operatorname{Aut}(N)$ defined in the obvious way instead?2011-09-16
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    @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: http://www.jstor.org/discover/10.2307/40656982?uid=3739832&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=56270277163) 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.2012-06-21
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    @user1729 The 3 questions you asked for general pairs (G,N) is an extremely hard question to address, such behaviour is for the most part only known in nice classes of groups. In particular, if G is d-generated and finite (d the size of a smallest generating set for G), and S=S_d(G) is the collection of all ordered generating d-tuples, the diagonal action of G on S is a free action, but not transitive in general. If it IS transitive, like for G=C_n, d=1 or G=Quaternions d=2, and d-generating tuple for G/N lifts to one for G! This is an old Thm of Gasch\"utz's. Then |Aut(G)|=|S| and .... well,2012-06-21

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