Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I haven't the slightest idea why (inner or outer) semi-direct group products are an interesting construction. I understand inner direct products, because you're just giving conditions for which a group can be considered the direct product of two of its subgroups, and I "get" direct products. They're a very simple construction, and showing that a group decomposes into that structure is a very strong statement.

But the outer semi-direct product construction seems totally arbitrary and bizarre. What's the intuition, here?

share|improve this question

3 Answers 3

up vote 6 down vote accepted

A very important question in group theory is the extension problem: given groups $H,N$, can you classify the groups $G$ that have $N$ as a normal subgroup with quotient $H$? In other words, groups that fit in a exact sequence $1 \to N \to G \to H \to 1$. Motivation from that comes from the classification of finite simple groups and the existence of composition series: if we could do all of that, and given these two other tools, we could classify all finite groups. That's very interesting!

As it turns out, the extension problem is very hard in general. The direct product $H \times N$ gives a "trivial" example of an extension of $H$ by $N$, but there's not a lot that can be said -- that's only one possible structure, there are many others.

There is one simplification though: split extensions. These are extensions such that there exists a section of the projection $p : G \to H$ (that is, a morphism $s : H \to G$ such that $ps = id$). This class of extension is exactly the class of semidirect products $N \rtimes H$. So semidirect products give a partial answer to the question of classifying group extensions.

I would also argue that the construction is not that far-fetched. I feel like the dihedral group is a great example. It's the set of isometries of a regular $n$-polygon. You've got rotations ($\mathbb{Z/nZ}$) and a symmetry with respect to a given line. Any element can be written as a product of a rotation and either the identity or the symmetry. But when you multiply two such elements, there's an interaction between the rotation and the symmetry and the rotation (it gets changed to the opposite rotation). If you think about it, it means that the dihedral group can be represented as a semidirect product $\mathbb{Z/nZ} \rtimes \mathbb{Z/2Z}$. So another possible answer is that the semidirect product is useful to construct interesting groups, for example the dihedral group.

share|improve this answer
So a split extension is one in which the quotient group $H$ has a set of "representatives" ($s(H)$ in your notation) which form a subgroup isomorphic to it? –  Jack M Feb 22 at 21:24
Yes, that's pretty much it. If you follow the steps of the proof you'll see that then, $G$ is the (inner) semidirect product $i(N) \rtimes s(H)$ (where $i : N \to G$ is the injection). –  Najib Idrissi Feb 22 at 21:27
Does it follow that every element of $G$ can be uniquely represented as a product of one of those representatives and an element of $N$? –  Jack M Feb 23 at 13:03

Direct products for matrix groups are pretty easy to understand: $$G\times H = \left\{ \begin{bmatrix} g & 0 \\ 0 & h \end{bmatrix} : g \in G, h \in H \right\}$$ They are just diagonal matrices. Multiplying elements of $G\times H$ can be done by looking at the $G$ and the $H$ parts separately. They have basically nothing to do with each other.

Sometimes though a group acts on something. For matrix groups, they tend to act on vector spaces. If $G$ is a matrix group and $V$ is a vector space for it (so that $V$ is a vector space of column vectors, and $gv \in V$ whenever $v \in V$) then we can form something similar to the direct product with upper triangular matrices: $$G\ltimes V = \left\{ \begin{bmatrix} g & v \\ 0 & 1 \end{bmatrix} : g \in G, v \in V \right\}$$

In this group the $G$ part can be done ignoring the $V$ part: $$\begin{bmatrix} g & v \\ 0 & 1 \end{bmatrix} \begin{bmatrix} h & w \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} gh & gw+v \\ 0 & 1\end{bmatrix}$$ but the $V$ part (the $w$) is altered by the way the $G$ part (the $g$) acts on $V$.

This sort of factorization happens automatically in some nice cases. For instance if $X$ is a finite group whose Sylow 2-subgroup $G$ is cyclic, or isomorphic to $C_4 \times C_2$, or many other common things, then $X$ automatically has such a $V$ (though it need not be a vector space, but the formulas for multiplication still work).

Also if a finite group has exactly one Sylow $p$-subgroup $V$, then $X$ has a $G$. Again $V$ need not be a vector space so $G$ need not be a matrix group, but the formulas still hold.

share|improve this answer

One important idea of a semidirect product $H \ltimes N$ is the group action of $H$ on $N$. The outer semidirect product just shows that any group action of $H$ on $N$ (from another point of view, any homomorphism $H \rightarrow \operatorname{Aut}(N)$) determines a group, in the "obvious way".

So if you want to study groups $G = HN$, where $N$ is normal in $G$ and $H \cap N = 1$, the construction of external semidirect products gives you all possible such groups (and you might want to study these, since there are many important examples of semidirect products, and they are common).

The definition of the operation in external semidirect product is very natural, in $G$ above we have

$$(h'n')(hn) = (h'h)((h^{-1}n'h)n)$$

and now here $n \mapsto h^{-1}nh$ is the group action. So given $H$, $N$ and a group action $\phi: H \rightarrow \operatorname{Aut}(N)$ (with $h \mapsto \phi_h$), to make the set of pairs $(h,n)$ into a semidirect product, we should define

$$(h',n') \cdot (h,n) = (h'h, \phi_h(n')n)$$

and this turns out to work.

share|improve this answer
What do you mean by "the obvious way" ? –  Jack M Feb 22 at 22:12
Perhaps that is badly phrased. But if you read further, I'm trying to explain why the definition is a bit obvious, and not arbitrary and bizarre. –  spin Feb 22 at 22:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.