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The goal of this exposition is to carefully explain the correspondence between the notions of external and internal semi-direct products of groups, as well as the connection between semi-direct products and short exact sequences.
Naturally, we start with the construction of semi-direct products.
Definition 1 Let $H$ and $Q$ be groups and let
 be a group homomorphism. The semi-direct product $H \semidirect_\theta Q$ is defined to be the group with underlying set $\{(h,q) \st h \in H,\ q \in Q\}$ and group operation $(h,q)(h',q') := (h \theta(q) h', qq')$ .
We leave it to the reader to check that $H \semidirect_\theta Q$ is really 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 notation whenever this map is clear from the context.
Set $G := H \semidirect Q$ . There exist canonical monomorphisms $H \lra G$ and $Q \lra G$ , given by \begin{eqnarray*} h \mapsto (h,1_Q),& \ & h \in H \\ q \mapsto (1_H,q),& \ & q \in Q \end{eqnarray*}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 $G := H \semidirect Q$ as above. Then:
Proof. Let $p: G \lra Q$ be the projection map defined by $p(h,q) = q$ . Then $p$ is a homomorphism with kernel $H$ . Therefore $H$ is a normal subgroup 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-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 $G$ .
- $HQ = G$ .
- $H \intersect Q = \{1_G\}$ .
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 $H \intersect Q = \{1_G\}$ . 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 $h'q'$ . Then $hq=h'q'$ so ${h'}^{-1}h = q' q^{-1} \in H \intersect Q = \{1_G\}$ . Therefore $h=h'$ and $q=q'$ . 
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 $G \iso H \semidirect_\theta Q$ where
 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 $\phi: H \semidirect Q \lra G$ given by $\phi(h,q) = hq$ is a bijection from $G$ to $H \semidirect Q$ . It only remains to show that this bijection is a homomorphism.
Given elements $(h,q)$ and $(h',q')$ in $H \semidirect Q$ , we have $$ \phi((h,q)(h',q')) = \phi((h \theta(q)(h'),qq')) = \phi(hqh'q^{-1},qq') = hqh'q' = \phi(h,q) \phi(h',q'). $$ Therefore $\phi$ is an isomorphism. 
Consider the external semi-direct product $G := H \semidirect_\theta Q$ with subgroups $H$ and $Q$ . We know from Theorem 5 that $G$ is isomorphic to the external semi-direct product $H \semidirect_{\theta'} Q$ , where we are temporarily writing $\theta'$ for the conjugation map $\theta'(q)(h) := qhq^{-1}$ of Theorem 5. But in fact the two maps $\theta$ and $\theta'$ are the same: $$ \theta'(q)(h) = (1_H,q) (h,1_Q) (1_H, q^{-1}) = (\theta(q)(h),1_Q) = \theta(q)(h).
$$ 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 An exact sequence of groups $$ 1 \lra H \stackrel{i}{\lra} G \stackrel{j}{\lra} Q \lra 1. $$ is split if there exists a homomorphism $k: Q \lra G$ such that $j \comp k$ is the identity map on $Q$ .
Theorem 7 Let $G$ , $H$ , and $Q$ be groups. Then $G$ is isomorphic to a semi-direct product $H \semidirect Q$ if and only if there exists a split exact sequence $$ 1 \lra H \stackrel{i}{\lra} G \stackrel{j}{\lra} Q \lra 1. $$
Proof. First suppose $G \iso H \semidirect Q$ . Let $i: H \lra G$ be the inclusion map $i(h) = (h,1_Q)$ and let $j: G \lra Q$ be the projection map $j(h,q) = q$ . Let the splitting map $k: Q \lra G$ 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 $k: Q \lra G$ 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 $gq^{-1} \in \Im i$ . 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 $k(q) \in \Im i = \ker j$ , so $q = j(k(q)) = 1_Q$ . Therefore $g = k(q) = k(1_Q) = 1_G$ , so $i(H) \intersect k(Q) = \{1_G\}$ .
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 $H \semidirect Q$ . 
Thus, not all normal subgroups $H \subset G$ give rise to an (internal) semi-direct product $G = H \semidirect G/H$ . More specifically, if $H$ is a normal subgroup of $G$ , we have the canonical exact sequence $$ 1 \lra H \lra G \lra G/H \lra 1. $$ We see that $G$ can be decomposed into $H \semidirect G/H$ as an internal semi-direct product if and only if the canonical exact sequence splits.
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