Semidirect Products

Twist two groups by an automorphism action to build non-abelian ones.

The idea

Given two groups $H$ and $K$, the direct product $H \times K$ combines them in the laziest way available: multiplication runs in each coordinate separately and the two halves never interact. That laziness costs you something. If $H$ and $K$ are abelian then so is $H \times K$, so direct products of abelian groups will never produce a non-abelian group, and abelian groups alone are a thin supply of examples.

To do better, let $K$ act on $H$. An action assigns to each $k \in K$ an automorphism $\varphi_{k}$ of $H$, compatibly with multiplication in $K$ — which is to say it is a homomorphism $\varphi \colon K \to \operatorname{Aut}(H)$. Now build a group on the same set of pairs $(h, k)$, except that when a $k_{1}$ on the left passes an $h_{2}$ on the right, it acts on it: $(h_{1}, k_{1})(h_{2}, k_{2}) = \bigl(h_{1}\,\varphi_{k_{1}}(h_{2}),\; k_{1} k_{2}\bigr).$ This is the semidirect product $H \rtimes_{\varphi} K$. Verifying associativity uses exactly the fact that $\varphi$ is a homomorphism, and nothing else.

The result has order $|H| \cdot |K|$ and contains a copy of $H$ as a normal subgroup together with a copy of $K$. What it does not have to be is abelian, even when both ingredients are — the action supplies the non-commutativity. That is how the classification of small groups gets most of its stock of examples: fix two cyclic groups, vary the action, and read off which groups appear.

Ways to work on it

• Walkthrough. The twisted multiplication rule, when it is a direct product, and recognizing S_3.
• Practice. Count the homomorphisms K → Aut(H) that define semidirect products.
• Hardest. Build a non-abelian group of order pq and decide when one exists.

Not sure where to start? Take the ten-question placement test.