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In the Finite Group book by Isaacs, the concept of complement is defined as follows: Given the normal subgroup of $G$, a subgroup $H \subseteq G$ is an even complement $N$ in $G$ if $NH = G$ and $N \cap H = 1$. If $H$ complements $N$ in $G$, obviously each conjugate of $H$ in $G$ also complements $N$. In General, however, a Normal subgroup of $G$ may have unconjugated complements. An example where the complements are not conjugated occurs in the Klein group. I'm trying to find another example

Exodd
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Sera
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  • What about considering the alternating group $A_n$ and various odd involutory permutations in the permutation group $S_n$? When $n\geq6$ there are some of these that are not conjugated. – AdLibitum Jan 10 '17 at 11:29

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Consider $G=C_p \times C_p$, the direct product of the cyclic group $C_p$ with itself, where $p$ is a prime number. Let $H$ be any subgroup of $G$ of order $p$. Then any subgroup of $G$ of order $p$ different from $H$ is a complement of $H$ in $G$, and all these subgroups are obviously not conjugated since $G$ is abelian.

Martino
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