How to "feel" semidirect products

Question

Sorry if the question makes no sense, but I've always had a problem "feeling" semidirect products though I understand them.

Unlike direct products, I can easily point it out when I'm working with a group. This happens when some of the elements seem independent of other elements. Like when I'm adding complex numbers, I can see the imaginary part go with the imaginary part, and the real with the real. Or when multiplying, the magnitude multiplies by the magnitude, and the angle adds with the angle.

But I am not able to develope an intuitive feeling to know when a group is a semidirect product of 2 subgroups. The most thing I've went to so far is to see some elements independent but interact in special cases. Like in the dihedral group, where rotations are rotations, and a reflection is a reflection unless there are 2 reflections which might affect a rotation. But other than that, pointing a semidirect product is not as obvious as direct products. Can someone help me quickly notice when a group is a semidirect product of 2 subgroups?

Answer 1

You are asking how to recognize a group as being a semidirect product, rather than how to build new groups using the semidirect product construction.

Fact: a group $G$ is a semidirect product of two subgroups $H$ and $K$ if

(i) $G = HK$ (each element of $G$ has the form $hk$ for some $h \in H$ and $k \in K$),

(ii) $H \cap K = \{1\}$ (that makes the representation $g = hk$ unique),

and

(iii) $H$ or $K$ is a normal subgroup (it doesn't matter which one is normal since $HK = KH$ when either subgroup is normal, e.g., if $K \lhd G$ then $kh = h(h^{-1}kh)$, and $h^{-1}kh \in K$).

See Section 4, especially Theorem 4.1, here.

Example: Let $G = D_n$ and $H = \langle r\rangle$ be the subgroup of rotations. This has index $2$, so it's normal in $G$. A reflection $s$ in $G$ is outside of $H$ and has order $2$, so $H$ and $K := \langle s\rangle$ satisfy (i), (ii), and (iii). Thus $G$ is a semidirect product of $H$ and $K$.

This won't help you show a group is not a semidirect product of any two nontrivial subgroups, e.g., $Q_8$ (and more generally the generalized quaternion groups of order $2^n$ for $n \geq 3$) is not a semidirect product of two nontrivial subgroups. But that's not what you're asking about.

Answer 2

If $H,K\le G\,,HK=G$ and $H\cap K=e$, and only one of $H$ and $K$ is normal, then $G\cong H\rtimes K$. I think $H$ would be the normal one here.

In general you need a homomorphism $\varphi: K\to\rm{Aut}(H)$ (i.e. a group action). This is used to effect the semi-direct product, denoted $H\rtimes _\varphi K$.

There's other equivalent statements.

Of course, if they're both normal, we just get $G\cong H×K$.

The semi-direct product is a good way to construct examples of non-abelian groups, for one thing.

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So for example. Say we take a couple cyclic groups, $\Bbb Z_n,\Bbb Z_m$. First, we can't necessarily do it (non-trivially). We need, for a non-trivial $\varphi: \Bbb Z_n\to\Bbb Z_m^×$, that $n/\mid\rm{ker}\varphi \mid\mid \varphi (m)$. In particular if $(n,\varphi (m))=1$, there's only the trivial one.

But if you can do it, in a non-trivial way, then it's easy to show, using the definition, that $\Bbb Z_m\rtimes _\varphi \Bbb Z_n$ will not be abelian.

For one set of examples, $\Bbb Z_n\rtimes \Bbb Z_2$ is $D_{2n}$, the dihedral group. Here the action is by element inversion.

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There's more to semi-direct products. Key is how the multiplication is defined.

Answer 3

To elaborate on Cpc's answer, think about $G$ as a group presentation: that is, a set of words in some alphabet, subject to various algebraic simplification rules. $G$ is a direct product $G=H\times K$ iff

1. $H$ and $K$ are subgroups.
2. Every word $w\in G$ has a canonical (unique) representation $$w=hk$$ with $h\in H$, $k\in K$
3. Every element of $H$ commutes with every element of $K$.

A semidirect product drops condition (3). (This connects to the definition of semidirect product as a quotient of the free product.)

Thus $D_n=\{1,\sigma,\dots,\sigma^n\}\rtimes\{1,\rho\}$ is a semidirect product, because every transformation $w\in D_n$ has a canonical representation as $\sigma^k\rho^j$. For an infinite example, $\mathrm{GL}(k^n)=\mathrm{SL}(k^n)\rtimes k^{\times}$, because given $M\in\mathrm{GK}(k^n)$, we can always "factor out" the determinant: $$M=Td\quad\quad\quad(d\in k^{\times},\det{\!(T)}=1)$$

You should think of semidirect products as split exact sequences: whenever you have a normal subgroup, look at the corresponding quotient. If there's a canonical (homomorphic) way to represent the equivalence classes in your original group, then you're looking at a semidirect product. (In the matrix group example, the canonical representation might not be so obvious — if $Z(\mathrm{GL}(k^n))$ denotes constant multiples of the identity matrix, then I'm using $\mathrm{GL}(k^n)/\mathrm{SL}(k^n)\cong k^{\times}\cong Z(\mathrm{GL}(k^n))$.)