# Complemented characteristic not implies left-transitively complemented normal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., complemented characteristic subgroup) need not satisfy the second subgroup property (i.e., left-transitively complemented normal subgroup)
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## Statement

### Statement with symbols

It is possible to have the following situation: groups $H \le K \le G$ such that $H$ is a complemented characteristic subgroup of $K$ and $K$ is a complemented normal subgroup of $G$, but $H$ is not a complemented normal subgroup of $G$.

## Facts used

1. Complemented normal satisfies intermediate subgroup condition

## Proof

### Example of the dihedral group of order sixteen

Further information: Dihedral group:D16

Let $G$ be a dihedral group of order $16$. In other words, $G$ is given by the presentation: $G = \langle a,x \mid a^8 = x^2 = e, xax = a^{-1} \rangle$.

Consider the subgroups: $H = \langle a^2 \rangle, \qquad K = \langle a^2, x \rangle$.

• $H$ is complemented characteristic in $K$: $K$ is a dihedral group of order eight, with $H$ as the unique cyclic normal subgroup of order four, hence $H$ is characteristic in $K$. Further, $\langle x \rangle$ is a complement to $H$ in $K$.
• $K$ is complemented normal in $G$: Since $K$ has index two, it is normal in $G$ (subgroup of index two is normal, or alternatively, nilpotent implies every maximal subgroup is normal). Further, the two-element subgroup $\langle ax \rangle$ is a permutable complement to $K$ in $G$.
• $H$ is not complemented normal in $G$: If $H$ were complemented normal in $G$, it would also (by fact (1)) be complemented normal in the intermediate subgroup $L = \langle a \rangle$, which is a cyclic group of order eight. But we know that $H$ is not complemented in $L$, since any element of $L$ not in $H$ generates $L$.