blurbs/grouptheory/semidirect-product

9. Complementary Subgroups

If a group $G$ contains subgroups $H$ and $K$ such that $G=H K$ and $H ∩ K=\{1\}$, then $H$ and $K$ are called complementary subgroups. 6Note that complementary subgroups are not complementary as subsets, e.g., in $ℝ^2$ the subgroups $ℝ×\{0\}$ and $\{0\} × ℝ$ are complementary but they certainly are not complementary subsets of $ℝ^2$. For a normal subgroup $H ◃ G$, is there always a complementary subgroup $K ⊂ G$ ? If so, we'd then have $G ≅ H ⋊ _φ K$
The answer is no!

Example 9.1. Suppose $G$ is a cyclic $p$-group with $|G|>p$. Every subgroup of $G$ is normal. Pick a subgroup $H ⊂ G$ with $1<|H|<|G|$. Suppose $G=H K$, so $|K|>1$. Then $H$ and $K$ have subgroups of order $p$ by Cauchy's theorem (or check this directly in cyclic $p$-groups). The cyclic group $G$ has at most one subgroup per size, so the subgroups of order $p$ in $H$ and $K$ are the same, and thus $H ∩ K ≠ \{1\}$. This contradiction shows $G$ is not a semidirect product of two proper subgroups.

Example 9.2. Let $G=Q_8=\{ ± 1, ± i, ± j, ± k\}$. There is only one subgroup of order 2, namely $\{ ± 1\}$. All subgroups of $Q_8$ are normal. Pick $H ◃ Q_8$ with $1<|H|<8$. If $Q_8=H K$, then $K ≠ \{1\}$ and by the same argument as in the previous example we have $H ∩ K ≠ \{1\}$. In particular, $Q_8$ is not a semidirect product of two proper subgroups.
When $G=H K$ with $H ◃ G$ and $H ∩ K=\{1\}$ such subgroups $K$ can be far from unique.

Example 9.3. For $n ≥ 3, S_n=A_n \cdot\{1, τ \}$ for every transposition $\tau$.

Example 9.4. For $n ≥ 3, D_n=⟨ r⟩ \cdot\left\{1, r^i s\right\}$ for every reflection $r^i s$ where $0 ≤ i ≤ n-1$