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
example where complements are not conjugate (in group theory)
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finite-groups
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0What 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. – 2017-01-10
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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.