### Affinity in Generality

Let V be a vector space. Consider the set of all affine transformations of V: An affine transformation of V is a map from V to itself which can be expressed as

$x \mapsto L(x) + b$,

for some invertible linear map (automorphism) L and some vector b in V. The set of all affine transformations of V forms a group under composition and is called the affine group of V. Note that the set of automorphisms of V is a subgroup of the affine group and also that the set of translations of V is also  subgroup of the affine group. Note that every element of the affine group can be expressed as a composition of an invertible linear map followed by a translation. Note also that the only affine transformation which is both an automorphism and a translation is the identity map. It is also quite easy to see that the translation group is a normal subgroup of the affine group. However, the automorphism group of V, which, since V is a vector space is known as the general linear group of V, is not.

Is the affine group the internal direct product of the translation group and the automorphism group? No, but it comes close. In the previous post, we saw that if G is the internal direct product of H and K, then not only can every element of G can be expressed (uniquely) in the form g = hk, but also the elements of H commute with the elements of K. This is the sense in which the two groups H and K do not interact with each other inside G. In the case of the affine group, there is interaction: The translation group of a vector space does not commute with its general linear group. This breakdown is evidenced by the fact that one of the two groups fails to be normal.

This is the second recent post on groups. In the first, I wrote about the concept of the direct product of groups. I outlined that there are two different kinds of direct product, external and internal. They are obviously in some sense the same thing, but I was also explicit about the differences which I think are important from a pedagogical point of view, if nothing else.

In this post, I will discuss what is known as a semi direct product of groups. As before, there are internal semidirect products and external semidirect products. However, this time, I will practice what I preached in the conclusion of the previous post and introduce the internal semidirect product first.

### Commutation

Suppose we have a group G with subgroups H and K such that

i) Every element g in G can be written as g=hk for h in H and k in K. And,

ii) The only element in both H and K is the identity, 1.

These two facts are true for the aforementioned affine group, but by working in this more general setup, let us take a close look at the difference between the way in which the affine group is built out the general linear group and the translation group on the one hand and the way in which direct products work on the other.

For elements x and y in G, write $x = h_1k_1$ and $y = h_2k_2$.  We’ll also write $[a,b] = aba^{-1}b^{-1}$. This is called the commutator of a and b.

Let’s naively see what happens if we try, in the most straightforward way possible, to express the product xy in the form hk. In a direct product, we would be able to swap the middle two terms on the right-hand side in the expression $xy = h_1k_1h_2k_2 ,$ to simply get $xy = h_1h_2k_1k_2.$ However, in general, how might we possibly be able to express xy in the form hk? Let’s look for the most basic approach:  If we crudely attempt to make the same swap which one can make in the direct product case, we get

$xy = h_1[k_1, h_2] h_2k_1k_2.$

Is this in the form hk? We have gained a commutator, which expresses the potential failure to commute, but, on inspection, there is some hope: The last two terms firmly belong to K. If we continue down this road, the question is, does $h_1 [k_1$,$h_2] h_2$ belong to H? Writing this expression out in full and cancelling terms where possible, we see that it is equal to $h_1k_1h_2k_1^{-1}$. Since, $h_1$ is in H, this expression belongs to H precisely when $k_1h_2k_1^{-1}$ does. Aha! In our case, this is true for all x and y precisely if H is normal in G.  We have answered the question: “What extra condition do we need so that the most basic method of putting xy in the form hk actually works?” Answer: We need H to be normal in G. The important thing to note is that no conclusions are drawn about K (we could have done things slightly differently and concluded that K must be normal, but the point is that one of them must to be). (Note that the two facts i) and ii) together are equivalent to the assertion that each element g in G can be expressed uniquely in the form hk, and morally this is what one wants to call a product of groups – see Zappa Szep Product for the more general case of this phenomena without either group necessarily being normal.)

In conclusion, what we’ve seen here is that the conjugation of H by K is the key: In a direct product, the expression $k_1h_2k_1^{-1}$ trivially belongs to H because the k’s commute with the h’s and it is therefore equal to $h_2$. In the slightly more general scenario we have analysed above, the expression is not necessarily equal to $h_2$. However, H is still normal and so is it is some other element of H.

### Theorisation

Let’s officially get the definition out of the way before moving on to an interpretation of the previous paragraph. If  the following three things hold:

i) Every element g in G can be written as g=hk for h in H and k in K.

ii) The only element in both H and K is the identity, 1.

iii) H is normal in G.

Then we say that G is the internal semidirect product of H by K. One writes $G = H \rtimes K$. If both K and H are normal in G, then, as we know, G is the internal direct product of H and K. So, the ‘semi’ really does mean half – one rather than both of the subgroups are normal.

Now, let’s think about the analysis we did in the previous section: An algebraist would know exactly how to interpret things. If G is the internal semidirect product of H by K, then each element k of K gives rise to an automorphism of H via conjugation:

$\varphi_k(h) = khk^{-1},$

This is an automorphism of H because H is normal. What we’ve actually defined is a homomorphism $\varphi$ from K to the automorphism group of H called the conjugation homomorphism ($\varphi(k) = \varphi_k$ is the conjugation automorphism of H defined above).  The punchline is that this homomorphism is trivial if and only if G is the direct product of H and K.

So, the interaction of H and K is expressed by the homomorphism $\varphi$. If there is no interaction, the product is direct and the homomorphism is trivial. If, on the other hand, there is interaction, the product is not direct, one of the two groups is not normal, they do not commute with each other, and the homomorphism we’ve just defined tells us exactly how the group structure deviates from that of a direct product: With $x = h_1k_1$ and $y = h_2k_2$ as before, we have

$xy = h_1\varphi_{k_1}(h_2)k_1k_2$.

To tie things up nicely, think about what happens when you conjugate the translation $x \mapsto x + b$ of a vector space V by an element L of GL(V).

### Externality

To finish off, let’s take a quick look at the external semidirect product. This was how I was first introduced to the idea. of semidirect products. It looked arbitrary and now, four years later, I have finally gotten over that initial disgust. These things take time.

In the previous section we focussed on the homomorphism between one of the groups in the product and the automorphism group of the other. This is indeed the core of the semidirect product. Suppose I have two groups H and K and a homomorphism $\varphi$ from K to Aut(H). Is there a group G containing H and K (or isomorphic copies thereof) as subgroups, with H normal, which is the internal semidirect product of said subgroups?

Let’s try to answer this question. The final displayed equation of the previous section will be our guide. We want something a bit like the external direct product of H and K but with some non-trivial interaction between the summands, governed by $\varphi$ in the manner we have discussed above. We therefore start with the Cartesian product H$\times$K and bestow it with a product given by:

$(h_1,k_1)\ast(h_2,k_2) = (h_1\varphi(k_1)(h_2),k_1k_2).$

It should by this stage be obvious that this is the right thing to do. The group which results is called the external semidirect product of H by K corresponding to $\varphi$. Suffice to say that the set of elements in H$\times$K of the form (h,1) form a normal subgroup isomorphic to H and those of the form (1,k) a subgroup isomorphic to K.  Pleasingly, H$\times$K is the internal semidirect product of these subgroups and the corresponding conjugation homomorphism agrees with $\varphi$ once these subgroups are identified with H and K.

I’ve been reading Alperin and Bell’s Groups and Representations.

You might enjoy, as I did, reading about Wreath Products, Holomorphs and Zappa-Szep products.