# Semidirect Products

Let H, N be two groups with H cyclic. Let $f,g:H \rightarrow Aut(N)$ be homomorphisms such that $N\rtimes _f H \cong N\rtimes _g H$. Then does $f(H)$ and $g(H)$ are conjugate in $Aut(N)$?

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This is a question from Dummit and Foote and I do not think it's suitable for mathoverflow. – T.B. Jan 22 '11 at 4:07
Oops, just saw that comment and deleted my answer. – Alex B. Jan 22 '11 at 4:14
Tin Bui, do you mean exercise 6, section 5.5, p. 184 (3rd edition)? Because this question is actually asking about the converse of that exercise (minus the assumption that $f$ and $g$ are injective if $H$ is infinite). It is still possible that it's not appropriate, but my brain is not in group theory mode right now and it's hard to tell. – Zev Chonoles Jan 22 '11 at 4:19
I voted to close, but I did so because I read the problem quickly and saw Tin Bui's comment, so I thought that he was looking for the converse of what he is looking for. I thus wish to reverse my vote to close, so this is a vote to keep the problem open. Before someone casts the final vote, someone else should comment and vote to cancel this vote to keep open. By the way, the answer to this question is no, there definitely exist counterexamples. However, I'm too tired to work them out right now, and surely someone else will post them by the time I wake up... – Andy Putman Jan 22 '11 at 7:46
In light of the above comments, I have undeleted my answer. @Tin the OP could have had his response within 20 minutes, had it not been for your first comment, so it's slightly disingenuous to now write "you can get better and faster responses elsewhere". – Alex B. Jan 22 '11 at 8:28

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.