Semidirect Products

Section 7.2 Semidirect Products

Subsection External Semidirect Products

“Don’t give up and always keep on believing in your product.”
―Niels van Deuren

We now discuss an important generalization for the direct product and a new method of constructing a new groups from the action of one group on another.

Suppose \(G\) is a group with subgroups \(H\mathrel{\unlhd}G\) and \(K\leq G\) such that \(H\cap K=\{e\}\text{.}\) Then we still have \(HK\leq G\text{;}\) let’s see what we would need the multiplication on the cartesian product \(H\times K\) to be in order for \(\theta: H \times K \to HK\) defined by \(\theta(h,k) = hk\) to still be a group homomorphism:

\begin{equation*} \theta(h_1, k_1) \theta(h_2, k_2) = h_1k_1h_2k_2 = h_1h'_2k_1k_2=\theta(h_1h_2',k_1k_2), \end{equation*}

where \(h'_2\in H\) is such that \(k_1h_2k_1^{-1}=h'_2\text{.}\)

This means that we would need to have \((h_1, k_1)(h_2, k_2)=(h_1h_2',k_1k_2)\) for \(\theta\) to be a homomorphism. This motivates the following definition.

Definition 7.11. External Semidirect Products.

Let \(H\) and \(K\) be groups and let \(\rho:K\to \operatorname{Aut}(H)\) be a homomorphism. The (external) semidirect product induced by \(\rho\) is the set \(H \times K\) with the binary operation defined by

\begin{equation*} (h,k)(h',k') = (h\rho(k)(h'),kk'). \end{equation*}

This group is denoted by \(H \rtimes_\rho K\text{.}\)

Discussion 7.1.

Decide amongst yourselves whether it should be spelled "semidirect", "semi-direct", "semi direct".

Before we prove that the construction above actually gives a group, let’s compute a few examples.

Example 7.12. Semidirect, Sans the Semi.

Given \(H\) and \(K\) we could always take \(\rho\) to be the trivial homomorphism, so that \(\rho(y)(x) = x\) for all \(y \in K\) and \(x \in H\text{.}\) Then \(K \rtimes_\rho H\) is just the usual direct product:

\begin{equation*} (y_1, x_1) (y_2, x_2) = (y_1 y_2, x_1 x_2). \end{equation*}

Example 7.13. .

Fix a group \(G\text{,}\) a normal subgroup \(H \unlhd G\) and a subgroup \(K \leq G\text{.}\) Then the function

\begin{equation*} \rho: K \to \operatorname{Aut}(H) \end{equation*}

given by \(\rho(x)(y) = xyx^{-1}\) for \(x \in K, y \in H\) is a homomorphism. Thus \(K\) acts on \(H\) via automorphisms.

Example 7.14. .

Let \(K = \langle x \rangle\) be cyclic of order \(2\) and \(H = \langle y \rangle\) be cyclic of order \(n\) for any \(n \geq 1\text{.}\) There is an automorphism of \(K\) that sends \(y\) to \(y^{-1}\text{.}\) This automorphism is its own inverse; i.e., it has order \(2\text{.}\) Therefore, by Proposition 2.27, there is a group homomorphism

\begin{equation*} \rho: K \to \operatorname{Aut}(H) \end{equation*}

with \(\rho(x)(y) = y^{-1}\text{.}\) We may thus form the group

\begin{equation*} G := H \rtimes_\rho K. \end{equation*}

The elements of \(G\) are \((y^i, x^j)\) for \(0 \leq i \leq n-1\) and \(0 \leq j \leq 1\text{,}\) in particular \(|G|=2n\text{.}\) Set

\begin{equation*} \tilde y = (y,e) \in G \text{ and } \tilde x = (e,x) \in G. \end{equation*}

Then

\begin{equation*} \tilde y^n = (y,e_K)^n = (y^n,e_K) = (e_H, e_K)= e_G, \end{equation*}

\begin{equation*} \tilde x^2 = (e_H,x)^2 = (e_H, x^2) = (e_H, e_K)= e_G, \end{equation*}

and

\begin{equation*} \tilde x \tilde y \tilde x \tilde y = (e_H,x)(y,e_K)(e_H, x)(y,e_K) = (\rho(x)(y),x)(\rho(x)(y),x) = (y^{-1},x)(y^{-1},x) = (y^{-1}y, e) = e_G. \end{equation*}

Looks familar!

Indeed, by the universal mapping property for \(D_{2n}\) we have a homomorphism

\begin{equation*} \theta: D_{2n} \to G \end{equation*}

such that \(\theta(r) = (y,e_K)\) and \(\theta(s) = (x,e_H)\text{.}\) Moreover, \(\theta\) is onto since

\begin{equation*} \theta(r^is^j)=(y^i, x^j) \text{ for all } 0 \leq i \leq n-1, 0 \leq j \leq 1 \end{equation*}

and since \(|D_{2n}|=|G|=2n\) it follows that \(\theta\) is a bijection. So the dihedral group is a semidirect product, in which the two component groups are cyclic of orders \(n\) and \(2\) respectively:

\begin{equation*} D_{2n} \cong \langle y \rangle \rtimes_\rho \langle x \rangle \end{equation*}

and \(\rho\) is the inversion homomorphism as described above.

Theorem 7.15. Semidirect Products are Groups.

If \(H\) and \(K\) are groups and \(\rho:K\to \operatorname{Aut}(H)\) is a homomorphism, then setting:

\(H \rtimes_\rho K\) is a group
\(H\cong H':=\{(h,e)\mid h\in H\}\mathrel{\unlhd}H \rtimes_\rho K\) and \(K\cong K':=\{(e,k)\mid k\in K\}\leq H \rtimes_\rho K\)
\((H \rtimes_\rho K )/ H' \cong K\text{.}\)

Proof.

The proof is straightforward but a bit messy. For associativity, note that

\begin{equation*} \begin{aligned} (y_1,x_1) \left( (y_2, x_2) (y_3, x_3) \right) &=(y_1,x_1) (y_2\rho(x_2)(y_3), x_2x_3)\\ &=(y_1\rho(x_1)\left(y_2\rho(x_2)(y_3)\right), x_1x_2x_3)\\ &=(y_1\rho(x_1)(y_2)(\rho(x_1)\circ \rho(x_2))(y_3), x_1x_2x_3)\\ &=(y_1\rho(x_1)(y_2)\rho(x_1x_2)(y_3), x_1x_2x_3)\\ \end{aligned} \end{equation*}

On the other hand

\begin{equation*} \begin{aligned} \left( (y_1,x_1) (y_2, x_2) \right) (y_3, x_3) & = (y_1 \rho(x_1)(y_2), x_1 x_2) (y_3, x_3) \\ & = (y_1 \rho(x_1)(y_2)\rho(x_1x_2)(y_3), x_1 x_2 x_3). \end{aligned} \end{equation*}

This gives associativity.

The fact that \((e,e)\) is a two-sided identity follows from the fact that \(\rho(e)(y) = \text{id}_H(y)=y\text{.}\)

Finally

\begin{equation*} \begin{aligned} (y,x) (\rho(x^{-1})(y^{-1}) ,x^{-1}) &= (y \rho(x)\left(\rho(x^{-1})(y^{-1}) \right), e) = (y (\rho(x)\circ\rho(x^{-1}))(y^{-1}), e) \\ &= (y \rho(e)(y^{-1}), e) = (yy^{-1},e)=(e,e), \end{aligned} \end{equation*}

and similarly

\begin{equation*} (\rho(x^{-1})(y^{-1}) ,x^{-1}) (y,x) = (e,e). \end{equation*}
Define a funtion

\begin{equation*} i: H \to H \rtimes_\rho K \end{equation*}

as \(i(y) = (y, e)\text{.}\) Then \(i\) is a homomorphism, since

\begin{equation*} i(y_1) i(y_2) = (y_1,e)(y_2,e) = (y_1\rho(e)(y_2) , ee) = (y_1 y_2, e) = i(y_1y_2). \end{equation*}

The map is clearly injective and hence its image is isomorphic to \(H\text{.}\) In fact, the image is normal since the second component of

\begin{equation*} (y,x) (y_2, e) (\rho(x^{-1})(y^{-1}), x^{-1}) \end{equation*}

is clearly \(e\text{.}\) Let us write this image as

\begin{equation*} H' := \{(y,e) \mid y \in H\} \unlhd H \rtimes_\rho K. \end{equation*}
The function

\begin{equation*} j: K \to H \rtimes_\rho K \end{equation*}

defined by \(j(x) = (e,x)\) is also an injective homomorphism and thus its image

\begin{equation*} K' := \{(e,x) \mid x \in H \} \leq H \rtimes_\rho K \end{equation*}

is isomorphic to \(K\text{.}\) \(K'\) is typically not normal, however. Finally, it is easy to see that \(H'K'= H \rtimes_\rho K\) and \(H' \cap K' = \{e\}\text{.}\) Putting this all together we have
- \(H' \unlhd H \rtimes_\rho K\text{,}\)
- \(K' \leq H \rtimes_\rho K\text{,}\)
- \(H' K' = H \rtimes_\rho K\text{,}\) and
- \(H'\cap K' = \{e\}\text{.}\)
Consider the projection onto the second factor \(\pi_2:H \rtimes_\rho K \to K\) given by \(\pi_2(y,x)=x\text{.}\) This is a goup homomorphism since the second component of \((y_1,x_1)(y_2,x_2)\) is \(x_1x_2\) and is surjective by definition. Now

\begin{equation*} \operatorname{Ker}(\pi_2)=\{(y,e_K)\mid y\in H\}=H'\cong H. \end{equation*}

By the First Isomorphism Theorem we conclude that \((H \rtimes_\rho K )/ H' \cong K\text{.}\)

Subsection Internal Semidirect Products

“I’ve often said there’s nothing better for the inside of a man than the outside of a horse.”
―Ronald Reagan

We can turn this around.

the semidirect product, not the Reagan quote.

Theorem 7.16. Recognition Theorem for Internal Semidirect Products.

For a group \(G\text{,}\) suppose we are given \(H\) and \(K\) so that

\(H \unlhd G\text{,}\)
\(K \leq G\text{,}\)
\(HK = G\text{,}\) and
\(H \cap K = \{e\}\text{.}\)

Let \(\rho: K \to \operatorname{Aut}(H)\) be the permutation representation of the action of \(K\) on \(H\) via automorphisms given by Conjugation in \(G\text{.}\) (This means that for any \(k\in K\) \(\rho(k)=c_k\text{,}\) where \(c_k\in \operatorname{Aut}(H)\) is the function \(c_k(h)=khk^{-1}\) for all \(h\in H\text{.}\)) Then the function

\begin{equation*} \theta: H \rtimes_\rho K \to G \end{equation*}

defined by \(\theta(y,x) = yx\) is an isomorphism of groups. Moreover, under this isomorphism, \(K\) corresponds to \(K'\) and \(H\) corresponds to \(H'\) (referring to the notation in Theorem 7.15 above).

Proof.

We have

\begin{equation*} \begin{aligned} \theta((y_1, x_1) (y_2, x_2)) & = \theta(y_1 c_{x_1}(y_2), x_1 x_2) \\ & = y_1x_1y_2x_1^{-1} x_1 x_2 \\ & = y_1x_1y_2x_2 = \theta(y_1, x_1) \theta(y_2, x_2) \end{aligned} \end{equation*}

and thus \(\theta\) is a homomorphism. It’s kernel is \(\{(y,x) \mid y = x^{-1}, y\in H, x\in K \}\text{,}\) which is just \(\{e\}\) since \(H \cap K = \{e\}\text{.}\) The image of \(\theta\) is clearly \(KH = G\text{.}\) This proves \(\theta\) is an isomorphism. It is obvious that \(\theta(K') = K\) and \(\theta(H')= H\text{.}\)

Definition 7.17. Internal Semidirect Products.

In this situation of the Theorem 7.16, we will say that \(G\) is the internal semi-direct product of \(H\) and \(K\text{.}\)

Example 7.18. .

Returning to \(D_{2n}\text{,}\) let \(H = \langle s \rangle\) and \(K = \langle r \rangle\text{.}\) Then \(H \leq G\text{,}\) \(K \unlhd G\text{,}\) \(HK = G\) and \(H \cap K =\{e\}\text{.}\) So, \(G\) is isomorphic to a semi-direct product, as we already showed.

Example 7.19. .

Let \(G = S_n\text{,}\) \(K = A_n\) and \(H = \langle (1 \, 2) \rangle\text{.}\) Then \(K \unlhd G\text{,}\) \(H \leq G\text{,}\) \(KH = G\) and \(K \cap H = \{e\}\text{.}\) It follows that

\begin{equation*} S_n \cong A_n \rtimes_\rho C_2 \end{equation*}

where \(C_2 = \langle x \rangle\) is cyclic of order \(2\) and the action \(\rho: C_2 \to \operatorname{Aut}(A_n)\) sends \(x\) to conjugation by \((1 \, 2)\text{.}\)

It is important to be aware that for a fixed pair of groups \(H\) and \(K\text{,}\) different actions of \(H\) on \(K\) via automorphisms can result in isomorphic semi-direct products. Indeed, determining when \(K \rtimes_{\rho} H \cong K \rtimes_{\rho'} H\) is in general a tricky business. The previous example shows this.

Example 7.20. .

Let \(G = S_n\) and \(K = A_n\) again, but this time take \(H' = \langle (1 \, 3) \rangle = (1 \,2 \, 3) \langle (1 \, 2) \rangle (1 \,2 \, 3)^{-1}\) (assuming \(n \geq 3\)). Then we get

\begin{equation*} S_n \cong A_n \rtimes_{\rho'} C_2 \end{equation*}

where \(C_2 = \langle x \rangle\) is cyclic of order \(2\) and the action \(\rho': C_2 \to \operatorname{Aut}(A_n)\) sends \(x\) to conjugation by \((1 \,3)\text{.}\)

The actions \(\rho\) and \(\rho'\) are not identical. For example, assuming \(n\geq 4\) we have

\begin{equation*} \rho(x)(1 \, 2) (1 \, 2) (3 \, 4) = (1 \, 2) (3 \, 4) \text { maps } 1\mapsto 2 \end{equation*}

and

\begin{equation*} \rho'(x)(1 \, 2) (3 \, 4) = (1 \, 3) (1 \, 2) (3 \, 4) (1 \, 3) \text { maps } 1\mapsto 4. \end{equation*}

Yet

\begin{equation*} A_n \rtimes_{\rho} H \cong A_n \rtimes_{\rho'} H' \end{equation*}

since each is isomorphic to \(S_n\text{.}\)

On HW 8 you will give a more conceptual reason for why these two semidirect products turned out to be isomorphic: it is because \(H\) and \(H'\) are conjugate in \(S_n\text{.}\) More generally, below is a criterion for a two semidirect products to be isomorphic.

Theorem 7.21. Isomorphisms of Semidirect Products.

Let \(K\) be a finite cyclic group and let \(H\) be an arbitrary group. Suppose that the images of \(\phi: K \to \operatorname{Aut}(H)\) and \(\theta: K \to \operatorname{Aut}(H)\) are conjugate subgroups of \(\operatorname{Aut}(H)\text{.}\) Then \(H \rtimes_\phi K \cong H \rtimes_\theta K\text{.}\)

Subsection Groups of Order \(pq\)

“I did not know I was in my prime until afterwards.”
―Mason Cooley

Theorem 7.22. Groups of Order \(6\).

Any group of order \(6\) is isomorphic either to \(C_6\) or to \(D_6\text{.}\)

Proof.

Let \(G\) be a group of order \(6\text{.}\) Cayley’s Theorem gives that there exist elements \(x,y\in G\) with \(|x|=2\) and \(|y|=3\text{.}\) Let \(K=\langle x \rangle\) and \(H=\langle y\rangle\text{.}\) Since \([G:H]=2\text{,}\) \(H\) is a normal subgroup of \(G\) and since \(H\cap K\) is a common subgroup of \(H\) and \(K\) Lagrange’s Theorem gives that \(|H\cap K|\mid \operatorname{gcd}(|H|,|K|)=1\text{.}\) Thus \(H\cap K=\{e\}\) and since \(|HK|=\frac{|H||K|}{|H\cap K|}=6=|G|\) we deduce that \(HK=G\text{.}\) Theorem 7.16 now gives that \(G\) is the internal semidirect product of \(H\) and \(K\text{.}\) More to the point, \(G\cong H\rtimes_\rho K\text{,}\) where \(\rho:K\to \operatorname{Aut}(H)\) gives the action of \(K\) on \(H\) by Conjugation.

We now analyze the possibilities for \(\rho\text{.}\) By Theorem 2.25, \(\operatorname{Aut}(H)\cong\operatorname{Aut}(\mathbb{Z}/3)\cong (\mathbb{Z}/3^\times, \cdot)=(\{\pm 1\},\cdot)\text{.}\) There are two possibilities for the image of \(\rho\text{:}\) either \(\operatorname{Im}(\rho)=\{\text{id}_H\}\) or \(\operatorname{Im}(\rho)=\operatorname{Aut}(H)\text{.}\)

If \(\operatorname{Im}(\rho)=\{\text{id}_H\}\text{,}\) then \(\rho(x)=c_x=\text{id}_H\) (which implies \(xy=yx\)) and \(H\rtimes_\rho K=H\times K\text{.}\) Therefore, in this case \(G\cong H\times K\cong C_3\times C_2\cong C_6\text{,}\) where the last isomorphism uses the Sunzi’s Remainder Theorem.

If \(\operatorname{Im}(\rho)=\operatorname{Aut}(H)\text{,}\) then \(\rho(x)\) is the map \(y^i\mapsto y^{-1}\) and by an earlier example for this \(\rho\) we have \(H\rtimes_\rho K\cong D_6\text{,}\) so \(G\cong D_6\text{.}\)

Finally, \(C_6\not \cong D_6\) because the former is abelian and the latter is not.

Let’s repeat the previous example for classifying groups of order \(p\cdot q\) with \(p, q\) distinct primes into isomorphism classes.

Theorem 7.23. Classification for Groups of Order \(pq\).

Let \(p<q\) be primes.

If \(p\nmid q-1\) there is a unique group of order \(pq\) up to isomorphism, namely \(C_{pq}\text{.}\)
If \(p\mid q-1\) there are exactly two groups of order \(pq\) up to isomorphism, namely \(C_{pq}\) and a non-abelian group.

Proof.

Let \(G\) be a group of order \(pq\) and let \(H, K\) be Sylow subgroups of order \(q\) and \(p\) respectively. We see that \(H\) is a normal subgroup using Theorem 5.16, since \([G:H]=p\) is the smallest prime that divides \(|G|\text{.}\)

Furthermore, since \(H\cap K\) is a subgroup of both \(H\) and \(K\) we have by Lagrange’s Theorem that \(|H\cap K|\mid \operatorname{gcd}(q,p)=1\text{,}\) so that \(H\cap K=\{e_G\}\text{.}\) From here it follows that

\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}=\frac{q\cdot p}{1}=pq=|G| \end{equation*}

and so \(HK=G\text{.}\) Theorem 7.16 now yields that

\begin{equation*} G \cong H \rtimes_\rho K \cong C_q\rtimes_\rho C_p \end{equation*}

for some homomorphism \(\rho: K \to \operatorname{Aut}(H)\text{,}\) equivalently \(\rho: C_p \to \operatorname{Aut}(C_q)\text{.}\) By Proposition 2.27 to give such a homomorphism \(\rho\) is equivalent to giving an element \(z\in \operatorname{Aut}(C_q)\) so that \(z^p=\text{id}\text{,}\) which will give \(\rho(y^j)=z^j\) for \(C_p=\langle y\rangle\text{.}\) Thus \(|z|\mid p\) yielding that either \(|z|=1\) or \(|z|=p\text{.}\)

Case 1: if \(|z|=1\) then \(\rho\) is the trivial homomorphism and thus \(G\cong C_q\times C_p\cong C_{pq}\text{.}\)

Case 2: if \(|z|=p\) then it must be the case by Lagrange that \(p\mid |\operatorname{Aut}(C_q)\text{.}\) By Corollary 2.26 we know that \(\operatorname{Aut}(C_q)\cong C_{p-1}\) is a cyclic group. Therefore we have that \(p\mid (q-1)\) if and only if there exists an element \(z\in \operatorname{Aut}(C_q)\) of order \(p\) by Theorem 2.23 (2). Moreover any such element \(z\) generates a subgroup of \(\operatorname{Aut}(C_q)\) of order \(p\text{.}\) Since there is a unique subgroup of a cyclic group of a given order by Theorem 2.23 (2) we see that the image of \(\rho\) is independent of the choice of \(z\text{.}\) Thus by Theorem 7.21 we conclude that all subgroups resulting from any choice of \(z\) of order \(p\) are isomorphic.

Moreover, from the explicit presentation of semidirect products of cyclic groups given in a homework problem we see that the resulting group is non-abelian; in particular it is not isomorphic to \(C_{pq}\text{.}\)

Summary

External Semidirect Products are defined using homomorphisms into automorphism groups.
²
See: Theorem 7.15
They are often used to construct non-abelian groups.
Internal Semidirect Products are similar to their direct counterparts, though only require that one of \(H\) or \(K\) be normal. They can be identified using the Recognition Theorem for Internal Semidirect Products.
Using semidirect products, we are able to provide a Classification for Groups of Order \(pq\).

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