Let $H$ and $N$ be two groups with $H$ cyclic. Let $f,g:H \rightarrow \mathrm{Aut}(N)$ be homomorphisms such that $N\rtimes _f H \cong N\rtimes _g H$. Then does that mean $f(H)$ and $g(H)$ are conjugate in $\mathrm{Aut}(N)$?
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2$\begingroup$ This is a question from Dummit and Foote and I do not think it's suitable for mathoverflow. $\endgroup$– T.B.Commented Jan 22, 2011 at 4:07
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1$\begingroup$ Dear Tin Bui, I didn't see that this question in Dummit-Foote. I am asking it for following reason. If $G$ has a normal subgroup $N$, and a subgroup $H$, with $N\cap H=1$, then $G$ is a semidirect product of $N$ by $H$. Consider simple case when $H$ is cyclic. Then To determine possible dofferent semidirect products of $N$ by $H$, we can look for only those homomorphisms from $H$ to $Aut(N)$, such that the images of $H$ under homomorphisms are not conjugate. But is this sufficient to get non-isomorphic semidirect products? I worked for small groups, but I couldn't find counterexample. $\endgroup$– SolubleCommented Jan 22, 2011 at 4:34
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$\begingroup$ If you consider $f$ and $g$ to be functors from the one-object groupoids determined by $H$ to $Grp$ (with the object part picking out $H$), then there is a natural transformation $f\Rightarrow g$ precisely when for each $h\in H$ there is an $a\in Aut(N)$ such that $af(h)a^{-1} = g(h)$. I'm pretty sure that a natural transformation from $f$ to $g$ induces an isomorphism between the extensions determined by $f$ and $g$. You may need your isomorphism to commute with the maps to $H$. in order to reverse this argument. $\endgroup$– David Roberts ♦Commented Jan 22, 2011 at 6:04
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$\begingroup$ @Rahul: I look back at the exercise and Zev was right. Sorry, although I do think you perhaps can get faster and more detailed response to this type of question at math.stackexchange.com as I did when I had tried it. $\endgroup$– T.B.Commented Jan 22, 2011 at 6:07
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1$\begingroup$ Many books of algebra/group theory, and many articles on Semi-direct product give the following theorem: "If H is cyclic and f,g:H --> Aut(N) are homomorphisms, such that f(H) and g(H) are conjugate subgroups of Aut(N), then these two homomorphisms give isomorphic semi-direct product of N by H. But I didn't see, whether its converse is true, in any book, article, even not as an exercise, or any counterexample. I worked for small groups, but couldn't get. $\endgroup$– SolubleCommented Jan 22, 2011 at 7:25
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The answer is no. You can easily have a situation where $f$ is the trivial map, while $g$ makes $H$ act through an inner automorphism of $N$, so that in both cases $N\rtimes H\cong N\times H$. For concreteness, let $N=D_8$ be the dihedral group of order 8, let $\sigma$ be a non-central involution in $N$, let $H=\langle h\rangle\cong C_2$, and define $g(h)(n)=\sigma^{-1} n\sigma\;\forall n\in N$. Then $G=N\rtimes_g H\cong N\times H$, since the subgroup $\langle h\sigma\rangle$ of $G$ is of order 2, intersects $N$ trivially, and commutes with it.
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1$\begingroup$ You beat me to it by several seconds... What if we ask the same question after passing to Out(N)? $\endgroup$ Commented Jan 22, 2011 at 8:27
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$\begingroup$ Oh, I guess you undeleted an old answer? $\endgroup$ Commented Jan 22, 2011 at 8:28
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$\begingroup$ @Dan That's right, sorry. See the comment thread on the question for the explanation of why I deleted. I do agree that the question is elementary, but I have tripped over semi-direct products more than once myself. While it's something I would expect any researcher to be able to figure out rather quickly, I wouldn't expect that any mathematicians must know the answer off the top of his head. $\endgroup$– Alex B.Commented Jan 22, 2011 at 8:31
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$\begingroup$ Yeah, I've actually posted this sort of example somewhere else on MO before. But I've never thought about the question of conjugacy of f(H) and g(H) in Out(N). $\endgroup$ Commented Jan 22, 2011 at 8:34
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$\begingroup$ For what it's worth "somewhere" is here: mathoverflow.net/questions/23456/… $\endgroup$ Commented Oct 11, 2016 at 4:56