# Retract not implies every permutable complement is normal

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## Statement

Suppose $G$ is a group and $H$ is a retract of $G$. In other words, $H$ has a normal complement in $G$: there exists a normal subgroup $N$ of $G$ such that $N \cap H$ is trivial and $NH = G$.

There may exist non-normal permutable complements to $H$ in $G$. In other words, there may exist a non-normal subgroup $K$ of $G$ such that $HK = G$ and $H \cap K$ is trivial.

## Proof

### Example

Further information: symmetric group:S4

Let $G$ be the symmetric group on the set $\{ 1,2,3,4 \}$, $H$ be the subgroup comprising permutations on $\{ 1,2,3 \}$. Then:

• $H$ has a normal complement, namely the subgroup $N = \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$.
• $H$ has a permutable complement that is not normal, namely the subgroup $K = \{ (), (1,2,3,4), (1,3)(2,4), (1,4,3,2)\}$.