# semidirect product of groups

###### Definition 1.

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

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

Set $G:=H\rtimes Q$. There exist canonical monomorphisms       $H\longrightarrow G$ and $Q\longrightarrow G$, given by

 $\displaystyle h\mapsto(h,1_{Q}),$ $\displaystyle h\in H$ $\displaystyle q\mapsto(1_{H},q),$ $\displaystyle q\in Q$

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 $G:=H\rtimes Q$ as above. Then:

• $H$$G$.

• $HQ=G$.

• $H\cap Q=\{1_{G}\}$.

###### Proof.

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–direct products.

###### 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–direct product (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 $G=HQ$ and $H\cap Q=\{1_{G}\}$. Then every element $g$ of $G$ can be written uniquely in the form $hq$, for $h\in H$ and $q\in Q$.

###### Proof.

Since $G=HQ$, we know that $g$ can be written as $hq$. Suppose it can also 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 the internal semi–direct product of $H$ and $Q$. Then $G\cong H\rtimes_{\theta}Q$ where $\theta:Q\longrightarrow\operatorname{Aut}(H)$ is given by

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

By Lemma 4, every element $g$ of $G$ can be written uniquely in the form $hq$, with $h\in H$ and $q\in Q$. Therefore, the map $\phi:H\rtimes Q\longrightarrow G$ given by $\phi(h,q)=hq$ is a bijection from $G$ to $H\rtimes Q$. It only remains to show that this bijection is a homomorphism.

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

 $\phi((h,q)(h^{\prime},q^{\prime}))=\phi((h\theta(q)(h^{\prime}),qq^{\prime}))=% \phi(hqh^{\prime}q^{-1},qq^{\prime})=hqh^{\prime}q^{\prime}=\phi(h,q)\phi(h^{% \prime},q^{\prime}).$

Consider the external semi–direct product $G:=H\rtimes_{\theta}Q$ with subgroups $H$ and $Q$. We know from Theorem 5 that $G$ is isomorphic to the external semi–direct product $H\rtimes_{\theta^{\prime}}Q$, where we are temporarily writing $\theta^{\prime}$ for the conjugation  map $\theta^{\prime}(q)(h):=qhq^{-1}$ of Theorem 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 products and split exact sequences of groups.

###### Definition 6.
 $1\longrightarrow H\lx@stackrel{{i}}{{\longrightarrow}}G% \lx@stackrel{{j}}{{\longrightarrow}}Q\longrightarrow 1.$

is split if there exists a homomorphism $k:Q\longrightarrow G$ 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 a semi–direct product $H\rtimes Q$ if and only if there exists a split exact sequence

 $1\longrightarrow H\lx@stackrel{{i}}{{\longrightarrow}}G% \lx@stackrel{{j}}{{\longrightarrow}}Q\longrightarrow 1.$
###### Proof.

First suppose $G\cong H\rtimes Q$. Let $i:H\longrightarrow G$ be the inclusion map $i(h)=(h,1_{Q})$ and let $j:G\longrightarrow Q$ be the projection map $j(h,q)=q$. Let the splitting map $k:Q\longrightarrow G$ be the inclusion map $k(q)=(1_{H},q)$. Then the sequence above is clearly split exact.

Now suppose we have the split exact sequence above. Let $k:Q\longrightarrow G$ 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 $j(gq^{-1})=j(g)j(k(j(g)))^{-1}=1_{Q}$, so $gq^{-1}\in\operatorname{Im}i$. Set $h:=gq^{-1}$. 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 $k(q)\in\operatorname{Im}i=\ker j$, 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 $H\rtimes Q$. ∎

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

 $1\longrightarrow H\longrightarrow G\longrightarrow G/H\longrightarrow 1.$

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

Title semidirect product of groups SemidirectProductOfGroups 2013-03-22 12:34:49 2013-03-22 12:34:49 djao (24) djao (24) 10 djao (24) Definition msc 20E22 semidirect product semi-direct product