What is the motivation for semidirect products?

Question

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?

Answer 1

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.

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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.

Answer 2

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.

Answer 3

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.

Answer 4

For outer semidirect products, there is the splitting lemma saying that the existence of a short exact sequence $1 \rightarrow N \rightarrow G \rightarrow H \rightarrow 1$ gives you that $G$ is isomorphic to a semidirect of $N$ and $H$.

An intuitive understanding of this is a generalisation of orthogonality, which is a property that requires a scalar product. In linear algebra, one probably have seen that an euclidean vector space can be decomposed into $W = V \oplus V^{\perp}$ with componentwise addition. The condition of exactness of this short exact sequence is $im({N \rightarrow G}) = ker({G \rightarrow H})$. Imaging embedding $V$ in $W$ and making a projection on $V^{\perp}$ will satisfy the exactness condition.

But we are looking at groups and their products! So there is no orthongonality available but we found a nice condition, namely the exactness of the short exact sequence, which captures the idea of decomposing the group into a product so that componentwise operations are still possible. The semidirect products look strange at first, but it is just as unusual as nonstandard scalar products at first. They still keep up the property of componentwise operation which is easy to handle.

Answer 5

When we encounter any group for the first time, a natural question is: how can I do computations in this group? How can I actually take $a$ and $b$ and determine the value of $ab$? This question really breaks down into two parts:

1. Come up with a suitable notation for elements of the group.
2. Determine how to, given the notation for $a$ and $b$, find the notation for $ab$.

For example, in kindergarten we all learn how to answer to these requirements using decimal notation for the group of integers. Note that having a good notation for the group is absolutely essential to what we really mean by "performing a calculation", since without specifying a notational scheme in which the answer is to be expressed, we could technically answer the question "what is $2+2$" by saying "it's the integer equal to $2+2$".

How we do this for a given group depends on how we define the group. Many times a group presents itself to us naturally via a generating set. We have in mind some set of elements of an existing group, and consider the subgroup generated by those elements. In that case, it's natural to look for notational schemes involving those generating elements.

A nice case is when the generating set is $A\cup B$, where $A$ and $B$ both generate subgroups $G$ and $H$ respectively, and every element of $A$ commutes with every element of $B$. Since an arbitrary element of $\langle A\cup B\rangle$ can be written as a product of $a_i$-s in $A$ and $b_i$-s in $B$, we can see that by commutativity we can shuffle all of the $b_i$-s over to the right hand side and end up with a product $a_1...a_nb_1...b_m=gh$ for $g\in G$, $h\in H$, so that we now have a nice simple notation for elements of our group: a product $gh$ with $g\in G$ and $h\in H$, or more abstractly just the pair $(g, h)$. Multiplication in this notation is simple: $(g_1h_1)(g_2h_2)=(g_1g_2)(h_1h_2)$ by commutativity, so what we basically have here is a direct product (modulo the caveat that the notation we have chosen may not be unique if $G$ and $H$ have a non-trivial intersection).

More often we don't have commutativity. However, sometimes although the identity $ab=ba$ fails, the weaker identity $ba=ab'$ holds: for $g\in G, h\in H$, there is always an $h'\in H$ such that $hg=gh'$.

Example. Consider the group of permutations on the set of all binary matrices of size $n\times n$. Let's consider all such transformations which can be obtained from the following moves:

1. Swap two rows.
2. Swap two columns.
3. Invert a row (turn $1$-s to $0$-s and vice versa)
4. Invert a column.

We can break this generating set apart into the set of "row and column swaps" $A$ and the set of "row and column inversions" $B$. Each generates some subgroup. You can check for yourself that if $a$ is a swap operation and $b$ is an inversion operation, then we have $ba=ab'$ for some other (potentially equal to $b$) inversion operation.

This means we can do something similar to what we did previously in our attempt to find a general notation for transformations obtained from our generators. In a general product of $a$-s and $b$-s, we can shuffle all the $b$-s over to right, but at the cost of transforming them into different $b$'s. Still, it remains true that a general element of $\langle A\cup B\rangle$ can be written as $gh$, for some $g\in G=\langle A\rangle$ and $h\in H=\langle B\rangle$. Now, how do we perform multiplications in that notation?

It turns out that this type of situation can always be modelled as a semi-direct product. Suppose that, as above, $ba=ab'$ for some $b'\in B$, for any $a\in A, b\in B$. We can write that $b'$ as $f_a(b)$, since its value certainly depends only on $a$ and $b$. Now what is this function $f_a$? Well, rearranging the equation:

$$f_a(b)=a^{-1}ba$$

So $f_a$ is always an automorphism on the ambient group! More than that, it's a conjugation, so that $a\to f_a$ is a homomorphism from $G=\langle A\rangle$ into the automorphism group.

Now return to the question posted above: how can we multiply together two elements written as $g_1h_1$ and $g_2h_2$? Well, starting from $g_1h_1g_2h_2$, what we want to do is move $h_1$ "through" $g_2$ to get an answer in the proper form. By writing down $h_1$ and $g_2$ as products of elements of $A$ and $B$ respectively, you can see that we have $h_1g_2=g_2f_{g_2}(h_1)$, and so, expressed in "pair notation":

$$(g_1, h_1)(g_2, h_2)=(g_1g_2, f_{g_2}(h_1)h_2)$$

Which is precisely the defining formula for a semi-direct product.