Semidirect product

In group theory, a semidirect product is a generalization of the direct product which expresses a group as a product of subgroups.

There are two ways to think of the construction. One is intrinsic: the condition that a given group $G$ is a semidirect product of two given subgroups $N$ and $H$ is equivalent to some special conditions on the subgroups. This concept is usually called an inner semidirect product: $G = N \rtimes H.$ Expressing a group $G$ as an inner semidirect product of two subgroups can help in the study of the group's behavior. It can also help classify $G$ up to isomorphism.

Another way to think of semidirect products is extrinsic: given two abstract groups $G_1,G_2$ with some specified relationship between them, given by a certain homomorphism $\phi$ (as defined below), one can construct a new group called the semidirect product (or outer semidirect product) $G = G_1 \rtimes_\phi G_2.$ In this way, one can build new, larger groups from smaller ones, with a construction that is more general and richer than a direct product.

The following theorem gives three equivalent definitions of an inner semidirect product.

Let $G$ be a group and let $N$ and $H$ be subgroups of $G,$ with $N$ normal. Then the following statements are equivalent:

1. $NH = G$ and $N \cap H = \{1\}.$

2. Every $g \in G$ can be written uniquely as $g = nh$ with $n \in N, h \in H.$

3. Define $\psi : H \to G/N$ in the natural way: $\psi(h) = {\overline{h}} = hN.$ Then $\psi$ is an isomorphism.

If these conditions hold, one writes $G = N \rtimes H$ and says that this expresses $G$ as an inner semidirect product of $N$ and $H.$

Note that $N$ must be normal for these statements to make sense.

The proof of the theorem is straightforward: see the last section of the wiki.

Let $G = S_3.$ Let $N$ be the normal subgroup of order 3 generated by a 3-cycle, and let $H$ be a subgroup of order 2 generated by a 2-cycle. Then $G = N \rtimes H$ (e.g. by condition (1)).

This example generalizes along two different lines: first, let $G = S_n,$ $N = A_n,$ $H$ a subgroup of order 2 generated by a 2-cycle. Then $G = N \rtimes H$ in exactly the same way.

Second, let $G = D_n,$ the dihedral group of order $2n.$ Then let $N$ be the index-2 subgroup generated by rotations of the $n$-gon, and let $H$ be an order-2 subgroup generated by a reflection. Then $G = N \rtimes H.$

In verifying that a group $G$ is a semi-direct product of two given subgroups, it is often useful to note that if $G$ is finite and $|N| \cdot |H| = |G|,$ then it suffices to check exactly one of the two conditions $NH = G$ and $N \cap H = \{1\}$ (e.g. by the Second Isomorphism Theorem). And further, if $|N|$ and $|H|$ are relatively prime and $|N| \cdot |H| = |G|,$ the condition $N \cap H = \{1\}$ is automatic by Lagrange's theorem, so both conditions hold automatically.

Let $G$ be a group of order $p^a q^b$ for distinct primes $p,q.$ Let $N$ be its Sylow $p$-subgroup and suppose that $N$ is normal. Show that $G = N \rtimes H$ for some subgroup $H.$

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Let $H$ be a Sylow $q$-subgroup. Then $|N| \cdot |H| = |G|$ and $|N|$ and $|H|$ are coprime, so the above remarks show immediately that $G = N \rtimes H.$

The outer semidirect product is a sort of generalization of the inner semidirect product. Instead of starting with $G$ and prescribed subgroups $N$ and $H,$ the outer semidirect product starts only with the abstract subgroups and constructs the semidirect product $G.$ The construction does require a notion of multiplication of elements of $N$ with elements of $H$; this extra structure is part of the data that is used to construct the group $G.$

Let $N$ and $H$ be groups, and suppose $\phi : H \to \text{Aut}(N)$ is a homomorphism, which sends elements $h \in H$ to automorphisms $\phi_h$ of $N.$ Then the group $G = N \rtimes_\phi H$ is defined as the set of ordered pairs $(n,h)$ with $n \in N, h \in H,$ and group law given by the formula $$(n,h) \cdot (n_1,h_1) = (n\phi_h(n_1),hh_1).$$

Remarks:

• The verification that $G$ is a group is nontrivial but straightforward.

• Note that the group $G$ depends very strongly on the homomorphism $\phi.$ Leaving $H$ and $N$ unchanged but changing $\phi$ will change the group $G$ (even up to isomorphism--two different $\phi$ can lead to two non-isomorphic groups $G$).

• Note that this group law is a sort of "twisted" version of the group law of the direct product. Another way to say this is that direct products are trivial examples of semidirect products:

If $N$ and $H$ are any groups, and $\phi : H \to \text{Aut}(N)$ is the trivial homomorphism, sending every $h\in H$ to the identity automorphism of $N,$ then $N \rtimes_\phi H = N \times H.$

• An outer semidirect product is an inner semidirect product. That is, given $N,$ $H,$ and a homomorphism $\phi : H \to \text{Aut}(N),$ the group $G = N \rtimes_\phi H$ has a normal subgroup ${\mathcal N} = \{ (n,1) : n \in N\}$ and a subgroup ${\mathcal H} = \{(1,h) : h \in H\},$ such that $G = {\mathcal N} \rtimes {\mathcal H}.$ (Of course ${\mathcal N} \cong N$ and ${\mathcal H} \cong H.$)

• An inner semidirect product is an outer semidirect product. That is, given a group $G$ with subgroups $N$ and $H$ such that $G = N \rtimes H,$ there is a natural homomorphism $\phi : H \to \text{Aut}(N)$ such that $G \cong N \rtimes_\phi H.$ That homomorphism is as follows: for $h \in H,$ define $\phi_h(n) = hnh^{-1}.$ Then $$n_1h_1n_2h_2 = n_1 (h_1n_2h_1^{-1}) h_1h_2 = (n_1\phi_{h_1}(n_2))(h_1h_2),$$ which looks like the group law for an outer semidirect product; in other words, the above computation shows that the bijection $G \to N \rtimes_\phi H$ defined by $nh \mapsto (n,h)$ is also a homomorphism.

Let $N = {\mathbb Z}_2 \times {\mathbb Z}_2$ and let $H = {\mathbb Z}_2.$ Consider the automorphism $\phi: H \to \text{Aut}(N)$ which sends the nontrivial element of $H$ to the homomorphism $(a,b) \mapsto (b,a).$ Then the group $N \rtimes_\phi H$ is a group of order 8. Which group of order 8 is it isomorphic to?

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The most illuminating way to show that $G$ is isomorphic to the dihedral group $D_4$ is to find subgroups isomorphic to $N$ and $H$ in $G$ such that the conjugation action of $H$ on $N$ is the same as the effect of $\phi.$ Let $\sigma$ be a 90-degree rotation and $\tau$ a reflection. Then $\sigma^4 = \tau^2 = 1$ and $\sigma\tau = \tau \sigma^{-1}.$

Let $N = \{ 1, \sigma^2, \tau, \sigma^2 \tau \}$ and let $H = \{1,\sigma \tau\}.$ Then $N \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ and $H \cong {\mathbb Z}_2,$ as desired. What does conjugation by the nontrivial element of $H$ do to $N$? It sends $\tau$ to $\sigma^2 \tau$ and vice versa, and fixes $1$ and $\sigma^2.$ Picking the isomorphism $N \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ that sends $\tau$ and $\sigma^2\tau$ to $(1,0)$ and $(0,1)$ respectively, we see that conjugation by $H$ has the effect of switching $(1,0)$ and $(0,1)$ and preserving $(0,0)$ and $(1,1),$ which is precisely the automorphism $\phi_h.$

Note that this gives an example of the phenomenon mentioned above, that different choices of $\phi$ can lead to nonisomorphic groups: if $\phi$ is chosen to be the trivial homomorphism, then $N \rtimes_\phi H \cong N \times H = {\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_2.$

Let $$G = \left\{ \begin{pmatrix} a&b\\0&1 \end{pmatrix} : a,b \in {\mathbb R}, a \ne 0 \right\},$$ with group operation given by matrix multiplication. Let $H$ be the subgroup of $G$ consisting of diagonal matrices $\begin{pmatrix} a&0 \\ 0 &1 \end{pmatrix},$ and let $N$ be the subgroup of $G$ consisting of matrices of the form $\begin{pmatrix} 1&b\\0&1 \end{pmatrix}.$ Then $N$ is normal, $NH = G,$ and $N \cap H = \{1\},$ so $G = N \rtimes H.$

Abstractly, $N \cong {\mathbb R}$ and $H \cong {\mathbb R}^*$, so $G$ is isomorphic to the (outer) semidirect product ${\mathbb R} \rtimes_\phi {\mathbb R}^*,$ where $\phi$ sends a nonzero real number $h$ to the automorphism of the additive group $\mathbb R$ given by multiplication by $h.$

Another way to view $G$ is as a group of affine transformations $x \mapsto ax+b.$ Then matrix multiplication in $G$ corresponds to composition of the associated affine transformations.

$(1) \Rightarrow (2):$ Since $NH = G,$ every $g$ can be written as $g=nh.$ The representation is unique because, if $nh = n_1h_1,$ then $n_1^{-1} n = h_1 h^{-1}.$ The left side is in $N$ and the right side is in $H$; since $N \cap H = \{1\}$ they must both be $1,$ so $n_1 = n$ and $h_1 = h.$

$(2) \Rightarrow (3):$ To see that $\psi$ is surjective, note that any coset $gN$ can be rewritten using (2) as $nhN = h(h^{-1}nh)N = hN,$ for some $n\in N, h \in H.$ (This is because $h^{-1}nh \in N,$ because $N$ is normal.) To see that $\psi$ is injective, suppose $h \in \text{ker}(\psi).$ Then $h \in N,$ but then $h = h \cdot 1 = 1 \cdot h$ expresses $h$ in two ways as a product of an element in $N$ and an element in $H.$ This representation is supposed to be unique, so $h$ must equal $1.$

$(3) \Rightarrow (1):$ If $h \in N \cap H,$ then $\psi(h) = hN = N$ is the identity of $G/N,$ so $h = 1$ because $\psi$ is injective. Now for any $g \in G,$ $gN = hN$ for some $h \in H$ because $\psi$ is surjective. So $g \in hN,$ so $g = hn$ for some $n \in N,$ so $g = (hnh^{-1})h \in NH.$ This shows that $NH = G.$