# Normal subgroup $K$ which has no complement.

Rotman's book on Group Theory claims that a normal subgroup $K\trianglelefteq G$ need not have a complement. I recall what is meant by complement.

Definition. Let $K$ be a (not necessarily normal) subgroup of a group $G$. Then, a subgroup $Q\leq G$ is a complement of $K$ in $G$ if $KQ=G$ and $K\cap Q=\{1\}$.

Thoughts: Firstly, I believe that the author refers to nontrivial normal subgroups because we can always take $K$ to be $G$ itself (since $G\trianglelefteq G$) and $Q=\{1\}$. Then clearly $K \cap Q=\{1\}$ and $KQ=G$.

The example I found for this claim is $\mathbb{Z}_{4}$. The only subgroups are the trivial ones and $\langle2_{4}\rangle$. Since the the group is cyclic, every subgroup is cyclic ($\implies$ abelian) and therefore normal. However if we take $K= \langle2_{4}\rangle$ and $Q=\{1\}$ then $KQ\neq\mathbb{Z}_{4}$.

Questions: Is my reasoning sound? Are there more interesting examples of normal subgroups which do not have a complement?

• Yes, your example is correct and it is the smallest such example. There are lots of examples, but opinions vary on what is interesting. – Derek Holt Nov 4 '17 at 13:06