# 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.

## Inner semidirect 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:

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

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

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 \(H.\)

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 \(G = N \rtimes H.\)

## Outer semidirect product

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 \(N\) and \(H\) be groups, let \(\phi : H \to \text{Aut}(N)\) be a homomorphism, and let \(G = N \rtimes_\phi H\) be the (outer) semidirect product. Suppose that \(G\) is abelian. Which of the following three statements must necessarily be true about \(H,N,\phi\)?

I. \(H\) and \(N\) are abelian

II. \(\phi\) is the trivial homomorphism

III. \(G\) is actually a direct product of groups

## Examples

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?

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.

## Proof of the theorem

\( (1) \Rightarrow (2):\) 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 \(NH = G.\)

**Cite as:**Semidirect product.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/semidirect-product/