ALG2_L10

Semidirect product

For two groups $H$ and $K$, the most basic construction of a group that contains copies of $H$ and $K$ as subgroups is the direct product $H\times K$. We can embed $H$ and $K$ into $H\times K$ “on the axes” by $h\mapsto (h,1)$ and $k\mapsto (1,k)$ for $h\in H$ and $k\in K$.

We can also use the direct product to decompose groups not initially constructed as a direct product. Given a group $G$ with subgroups $H$ and $K$, to recognize whether $G$ can be written as the direct product of $H$ and $K$, we first observe some properties of the embeddings of $H$ and $K$ into their direct product $H\times K$:

• they generate $H\times K$: $(h,k)=(h,1)(1,k)$,
• they intersect trivially: $(h,1)=(1,k)⟹h=1,k=1$,
• they commute element-wise: $(h,1)(1,k)=(1,k)(h,1)$.
  These properties can be turned around to craft a recognition theorem for a group $G$ to look like a direct product of two subgroups $H$ and $K$, as we did previously:

Theorem 1(Theorem).

Let $G$ be a group with subgroups $H$ and $K$ where

1. $G=HK$
2. $H\cap K=\{1\}$ in $G$,
3. $hk=kh$ for all $h\in H$ and $k\in K$.

Then the map $H\times K\to G$ by $(h,k)\mapsto hk$ is an isomorphism.

Note that in place of the third condition, we required $H◃G$ and $K◃G$ here. However, $H◃G,K◃G,H\cap K=\{1\}⟹hk=kh\forall h\in H,k\in K$, so we’re good.

There’s another way to construct a group using two groups $H$ and $K$, called the semidirect product. Interesting features include:

1. It may be nonabelian even if $H$ and $K$ are abelian (note that $H\times K$ is abelian iff $H$ and $K$ are abelian), and
2. there may be multiple nonisomorphic semidirect products using the same two groups.

Definition 2(Definition).

Given any two groups $H$ and $K$ and a group homomorphism $\varphi :K\to \text{Aut}(H)$, we can construct a new group $H{⋊}_{\varphi }K$, called the semidirect product of $H$ and $K$ with respect to $\varphi$, defined as follows:

1. As a set, $H{⋊}_{\varphi }K$ is the same as $H\times K$.
2. $(h_1,k_1)\circ (h_2,k_2)\equiv (h_1{\varphi }_{k_1}(h_2),k_1{k}_2)$.

Recognizing semidirect products

Theorem 3(Theorem).

Let $G$ be a group with subgroups $H$ and $K$ such that

1. $G=HK$,
2. $H\cap K=\{1\}$, and
3. $H◃G$.

Let $\phi :K\to \text{Aut}(H)$ be conjugation: ${\phi }_k(h)=kh{k}^{-1}$. Then, $\phi$ is a homomorphism and the map $H{⋊}_{\phi }K\to G$ by $(h,k)\mapsto hk$ is an isomorphism.