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Suppose that we have a group $A$ containing a normal subgroup $G$ and two complements $H$ and $K$.

Symbolically, $H, K \leq A$; $G \unlhd A$; $GH = GK =A$; $G \cap H = G \cap K = \{1\}$; so that $A \cong G \rtimes H$ and $A \cong G \rtimes K$.

Can I say that there is an automorphism of $A$ sending $H$ to $K$? Could this automorphism also fix $G$? What if $G$ is furthermore supposed characteristic in $A$? I can't see a simple way to prove it.

Thanks in advance.

Kyle Miller
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frafour
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  • First, you don't need to assume $H$ and $K$ are isomorphic. A complement of $G$ in $A$ must be isomorphic to $A/G$, so all complements are isomorphic. – verret Oct 16 '17 at 18:32
  • Second, your question is very strange. In your case, we have $G\rtimes H=A=G\rtimes K$, (equality, not isomorphism!), so we can take the isomorphism to be the identity on $A$, for example, which might not send $H$ to $K$. – verret Oct 16 '17 at 18:34
  • You are right, I edited the question – frafour Oct 16 '17 at 19:03

1 Answers1

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The answer is no. Let $A = G \times H$ with $G = S_3$ and $H =C_2$. So $H \le Z(A)$, and we can take $K$ to be a noncentral complement of $G$ in $A$.

The answer is still no if $G$ is characteristic in $A$. Take $G=A_5$ and $H=C_2$.

The answer is yes when $G$ is abelian, because then $H$ and $K$ must induce the same action $\phi$ on $G$.

Derek Holt
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    I was about to post the same example, see also here. – Mikko Korhonen Oct 16 '17 at 19:15
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    So more generally if $G$ is complete (trivial center and all automorphisms are inner), take a semidirect product $A = G \rtimes H$ with $H$ acting non-trivially on $G$. Then $G$ has both non-normal and normal complements in $A$, hence these complements are not conjugate under an automorphism. – Mikko Korhonen Oct 16 '17 at 19:26