semidirect product of groups

The goal of this exposition is to carefully explain the correspondencebetween the notions of external and internal semi–direct products ofgroups, as well as the connection between semi–direct products andshort exact sequences.

Naturally, we start with the construction of semi–direct products.

Definition 1.

Let $H$ and $Q$ be groups and let $\theta :Q⟶\mathrm{Aut}(H)$ be a grouphomomorphism. The semi–direct product $H{⋊}_{\theta }Q$is defined to be the group with underlying set $\{(h,q)\mid h\in H,q\in Q\}$ and group operation $(h,q)(h^{\prime },q^{\prime }):=(h\theta (q)h^{\prime },q{q}^{\prime })$.

We leave it to the reader to check that $H{⋊}_{\theta }Q$ isreally a group. It helps to know that the inverse of $(h,q)$ is$(\theta (q^{-1})(h^{-1}),q^{-1})$.

For the remainder of this article, we omit $\theta$ from the notationwhenever this map is clear from the context.

Set $G:=H⋊Q$. There exist canonical monomorphisms $H⟶G$ and $Q⟶G$, given by

$h\mapsto (h,1_Q),$$h\in H$

$q\mapsto (1_H,q),$$q\in Q$

where $1_H$ (resp. $1_Q$) is the identity element of $H$(resp. $Q$). These monomorphisms are so natural that we will treat $H$and $Q$ as subgroups of $G$ under these inclusions.

Theorem 2.

Let $G\mathrm{:=}H\mathrm{⋊}Q$ as above. Then:

• $H$ is a normal subgroup of $G$.

• $HQ=G$.

• $H\cap Q=\{1_G\}$.

Proof.

Let $p:G⟶Q$ be the projection map defined by $p(h,q)=q$. Then$p$ is a homomorphism with kernel $H$. Therefore $H$ is a normalsubgroup of $G$.

Every $(h,q)\in G$ can be written as $(h,1_Q)(1_H,q)$. Therefore $HQ=G$.

Finally, it is evident that $(1_H,1_Q)$ is the only element of $G$that is of the form $(h,1_Q)$ for $h\in H$ and $(1_H,q)$ for $q\in Q$.∎

This result motivates the definition of internal semi–directproducts.

Definition 3.

Let $G$ be a group with subgroups $H$ and $Q$. We say $G$ is the internal semi–direct product of $H$ and $Q$ if:

• $H$ is a normal subgroup of $G$.

• $HQ=G$.

• $H\cap Q=\{1_G\}$.

We know an external semi–direct product is an internal semi–directproduct (Theorem 2). Now we prove a converse (Theorem 5), namely, that an internal semi–direct product is an external semi–direct product.

Lemma 4.

Let $G$ be a group with subgroups $H$ and $Q$. Suppose $G\mathrm{=}H\mathit{}Q$ and $H\mathrm{\cap }Q\mathrm{=}\mathrm{\{}{\mathrm{1}}_G\mathrm{\}}$. Then every element $g$ of $G$ can be writtenuniquely in the form $h\mathit{}q$, for $h\mathrm{\in }H$ and $q\mathrm{\in }Q$.

Proof.

Since $G=HQ$, we know that $g$ can be written as $hq$. Suppose it canalso be written as $h^{\prime }{q}^{\prime }$. Then $hq=h^{\prime }{q}^{\prime }$ so $h^{\prime }{}^{-1}h=q^{\prime }{q}^{-1}\in H\cap Q=\{1_G\}$. Therefore $h=h^{\prime }$ and $q=q^{\prime }$.∎

Theorem 5.

Suppose $G$ is a group with subgroups $H$ and $Q$, and $G$ is theinternal semi–direct product of $H$ and $Q$. Then $G\mathrm{\cong }H{\mathrm{⋊}}_{\theta }Q$ where $\theta \mathrm{:}Q\mathrm{⟶}\mathrm{Aut}\mathit{}\mathrm{(}H\mathrm{)}$ is given by

$$\theta (q)(h):=qh{q}^{-1},q\in Q,h\in H.$$

Proof.

By Lemma 4, every element $g$ of $G$ can be writtenuniquely in the form $hq$, with $h\in H$ and $q\in Q$. Therefore,the map $\varphi :H⋊Q⟶G$ given by $\varphi (h,q)=hq$ is abijection from $G$ to $H⋊Q$. It only remains to show thatthis bijection is a homomorphism.

Given elements $(h,q)$ and $(h^{\prime },q^{\prime })$ in $H⋊Q$, we have

$$\varphi ((h,q)(h^{\prime },q^{\prime }))=\varphi ((h\theta (q)(h^{\prime }),q{q}^{\prime }))=\varphi (hq{h}^{\prime }{q}^{-1},q{q}^{\prime })=hq{h}^{\prime }{q}^{\prime }=\varphi (h,q)\varphi (h^{\prime },q^{\prime }).$$

Therefore $\varphi$ is an isomorphism.∎

Consider the external semi–direct product $G:=H{⋊}_{\theta }Q$ with subgroups $H$ and $Q$. We know from Theorem 5that $G$ is isomorphic to the external semi–direct product $H{⋊}_{{\theta }^{\prime }}Q$, where we are temporarily writing ${\theta }^{\prime }$for the conjugation map ${\theta }^{\prime }(q)(h):=qh{q}^{-1}$ ofTheorem 5. But in fact the two maps $\theta$ and${\theta }^{\prime }$ are the same:

$${\theta }^{\prime }(q)(h)=(1_H,q)(h,1_Q)(1_H,q^{-1})=(\theta (q)(h),1_Q)=\theta (q)(h).$$

In summary, one may use Theorems 2 and 5 to pass freely between the notions of internal semi–direct product and external semi–direct product.

Finally, we discuss the correspondence between semi–direct productsand split exact sequences of groups.

Definition 6.

An exact sequence of groups

$$1⟶H\stackrel{i}{⟶}G\stackrel{j}{⟶}Q⟶1.$$

is split if there exists a homomorphism $k:Q⟶G$ such that$j\circ k$ is the identity map on $Q$.

Theorem 7.

Let $G$, $H$, and $Q$ be groups. Then $G$ is isomorphic to asemi–direct product $H\mathrm{⋊}Q$ if and only if there exists asplit exact sequence

$$1⟶H\stackrel{i}{⟶}G\stackrel{j}{⟶}Q⟶1.$$

Proof.

First suppose $G\cong H⋊Q$. Let $i:H⟶G$ be theinclusion map $i(h)=(h,1_Q)$ and let $j:G⟶Q$ be theprojection map $j(h,q)=q$. Let the splitting map $k:Q⟶G$ bethe inclusion map $k(q)=(1_H,q)$. Then the sequence above is clearlysplit exact.

Now suppose we have the split exact sequence above. Let $k:Q⟶G$be the splitting map. Then:

• $i(H)=\ker j$, so $i(H)$ is normal in $G$.

• For any $g\in G$, set $q:=k(j(g))$. Then $j(g{q}^{-1})=j(g)j{(k(j(g)))}^{-1}=1_Q$, so $g{q}^{-1}\in \mathrm{Im}i$. Set $h:=g{q}^{-1}$. Then $g=hq$. Therefore $G=i(H)k(Q)$.

• Suppose $g\in G$ is in both $i(H)$ and $k(Q)$. Write $g=k(q)$. Then $k(q)\in \mathrm{Im}i=\ker j$, so $q=j(k(q))=1_Q$. Therefore $g=k(q)=k(1_Q)=1_G$, so $i(H)\cap k(Q)=\{1_G\}$.

This proves that $G$ is the internal semi–direct product of $i(H)$and $k(Q)$. These are isomorphic to $H$ and $Q$,respectively. Therefore $G$ is isomorphic to a semi–direct product $H⋊Q$.∎

Thus, not all normal subgroups $H\subset G$ give rise to an (internal)semi–direct product $G=H⋊G/H$. More specifically, if$H$ is a normal subgroup of $G$, we have the canonical exact sequence

$$1⟶H⟶G⟶G/H⟶1.$$

We see that $G$ can be decomposed into $H⋊G/H$ as an internal semi–direct product if and only if the canonical exact sequence splits.