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 $0 ≤ i ≤ n-1$