Semi-Direct Product

Key Definitions

Definition.
Consider a short exact sequence of groups

\[1 \to N \xrightarrow{\phi_1} G \xrightarrow{\phi_2} G_2 \to 1.\]

We say that the short exact sequence splits if one of the following equivalent conditions holds:

  1. There exists a homomorphism $\psi: G_2 \to G$ such that $\phi_2 \circ \psi = \mathrm{id}_{G_2}$.
  2. There exists a subgroup $H \leq G$ such that:
    • $H \cap N = {e}$, and
    • $HN = G$.

    In this case, $H \cong G_2 \cong G/N$.

Definition / Theorem.
Let $N$ and $H$ be groups, and let $f \in \operatorname{Hom}(H, \operatorname{Aut}(N))$. The semi-direct product $H \ltimes_f N$ is a group with underlying set $H \times N$ and multiplication

\[(h_1, n_1) \cdot (h_2, n_2) = \big(h_1 h_2, \ f(h_2^{-1})(n_1) \, n_2\big).\]

Key Theorems

Theorem (Schur–Zassenhaus).
A short exact sequence

\[1 \to N \to G \to H \to 1\]

splits if $\gcd(|N|, |H|) = 1$.

Theorem.
Let

\[1 \to N \to G \to H \to 1\]

be a splitting short exact sequence. Then there exists $f \in \operatorname{Hom}(H, \operatorname{Aut}(N))$ such that

\[G \cong H \ltimes_f N.\]

Remark.
In the splitting case, we can replace $N$ and $H$ by isomorphic copies inside $G$ (viewing them as subgroups).
Every element $g \in G$ can be written uniquely as $h n$ with $h \in H$ and $n \in N$.
This gives a bijection $G \leftrightarrow H \times N$ via $h n \mapsto (h, n)$. Then, we use group structure on $G$ to define a group structure on $H\times N$.

The multiplication in $G$ satisfies

\[h_1 n_1 \, h_2 n_2 = h_1 h_2 \, (h_2^{-1} n_1 h_2) \, n_2.\]

Since $N \trianglelefteq G$, we have $h_2^{-1} n_1 h_2 \in N$. Defining $f: H \to \operatorname{Aut}(N)$ by $f(h) = c(h)$ (conjugation by $h$) yields the desired formula for the semi-direct product structure.

Lemmas.
Let $H$ and $N$ be groups and $f \in \operatorname{Hom}(H, \operatorname{Aut}(N))$.

  1. If $f$ is the trivial homomorphism (sending every $h \in H$ to $\mathrm{id}_N$), then $H \ltimes_f N \cong H \times N$ (the direct product).
  2. If $f$ is non-trivial, then $H \ltimes_f N$ is non-abelian.

Exercises

Easy:

  1. (UCI 2025 June Algebra Qual) Give an example of a semi-direct product of two cyclic groups of odd order that is not abelian. (See Medium problem 2.)
  2. (UCI 2024 June Algebra Qual) Prove that every group of order $15$ is cyclic. (See Medium problem 1.)
  3. (UCI 2024 June Algebra Qual) Determine whether every group of order $21$ is cyclic. (See Medium problem 2.)

Medium:

  1. Let $p < q$ be primes with $p \nmid (q-1)$. Show that every group of order $pq$ is cyclic.
  2. Let $p < q$ be primes with $p \mid (q-1)$. Show that there exists a non-abelian group of order $pq$.
  3. Let $G$ be a group of order $2p$ where $p$ is an odd prime. Show that $G$ is either cyclic or dihedral.