# Dihedral Groups and Automorphisms, Part 2

Welcome back readers! In the last post, Dihedral Groups and Automorphisms, Part 1 we introduced the dihedral group. To briefly recap, the dihedral group $D_n$ of order $2n$ for $n\geq 3$ is the symmetry group of the regular Euclidean $n$-gon. Any dihedral group is generated by a reflection and a certain rotation. Moreover, in Part 1 we gave two other descriptions of the dihedral group $D_n$. The first is the presentation

$\langle r, s | r^n, s^2, sr = r^{-1}s\rangle.$

We also discovered that if we consider the cyclic group $C_n$ as a $C_2 = \{ 1, \sigma\}$ module via $\sigma*k = -k$, then $D_n$ is isomorphic to a semidirect product: $D_n\cong C_n\rtimes C_2$, which was the second description.

Now here in Part 2, we are going to learn something new about the dihedral group when $n$ is even: in this case, $D_n$ has an outer automorphism. But in order to prove this, we will introduce group cohomology!

I learned this application from Charles Weibel's book, "An Introduction to Homological Algebra", which I very much recommend.

### What is Group Cohomology?

First, I must say that a proper and civilized introduction to group cohomology would take much longer than a single post, so unfortunately we will only see a faint glimmer of it here, but the interested reader should check out some of the references at the end of this post.

What we will need for our purposes is a good knowledge of the first cohomology group $H^1(G,A)$ defined for a group $G$ and a $G$-module $A$. But what is this object? In fact it is an abelian group, and we shall see a few ways to define it. The first definition is this: consider first a derivation. A derivation $D:G\to A$ is a map that satisfies $D(gh) = D(g)h + D(h)$ for all $g,h\in G$. That's it. The set of all derivations form an abelian group under addition, and we will call this set $\mathrm{Der}(G,A)$. Furthermore, if $a\in A$ then the map $D_a(g) = ga – a$ is a derivation. Any derivation equal to $D_a$ for some $a\in A$ is called a principal derivation. The principal derivations form a subgroup of $\mathrm{Der}(G,A)$, and we shall denote the set of principal derivations by $\mathrm{PDer}(G,A)$.

We now define $H^1(G,A) = \mathrm{Der}(G,A)/\mathrm{PDer}(G,A)$. Lovely isn't it? Only, this is hardly motivated, and certainly not fun to calculate. That is why this is not really the correct definition of $H^1$ at all. And besides, we're trying to do cohomology here, so how can we define thing like $H^2$ and homology?

### What is Group Cohomology, Really?

Now we will give a more systematic definition, which although might sound complicated at first, actually makes life much simpler. Also, the reader can even skip this section and come back to it later.

A more natural definition is as follows: if $A$ is our $G$ module, then consider the invariant functor $-^G:A\mapsto A^G = \{ a\in A : \forall g\in G(ga = a)\}$. It is an additive functor from $G$-modules to abelian groups. This functor is left-exact, and so if we choose an injective resolution $A\to I$ then apply the $-^G$ functor and take the cohomology of the resulting complex, we get $H^0 = A^G$. Now, we just define $H^i(G,A) = H^i(I^G)$; here $I^G$ denotes the functor $-^G$ applied to the injective resolution.

It turns out that this is actually well-defined: the cohomology does not depend on the choice of injective resolution. This follows from a nifty trick called the comparison lemma for injective resolutions. Now the $H^i$ are actually the so-called right-derived functors of $-^G$. What this means is that we all of a sudden have a wealth of powerful theorems we can use to compute these cohomology groups. It turns out that the functor $-^G$ is a right-adjoint to the functor which sends an abelian group $M$ to the trivial $G$-module in the category of $G$-modules. This functor also has an additive left-adjoint called the coinvariant functor which we can define homology with, but we will not need this here.

Moreover, using the bar resolution, a nice canonical free resolution of $\mathbb{Z}$ that exists for any group $G$, it is not hard to prove that this new definition in the $H^1(G,A)$ case coincides with the derivation/principal derivation definition.

If all that was a bit complicated, it should be clarified a bit by the calculation a bit further down.

Consider a group $G$ and a $G$-module $A$. As we have seen (refer to Part 1 for the definitions), we can form the semidirect product $A\rtimes G$. Let $\varphi:A\rtimes G\to A\rtimes G$ be an automorphism of groups. We say that $\varphi$ is a stabilizing automorphism if $\varphi(a,1) = (a,1)$ for all $a\in A$ and the induced map on $(A\rtimes G)/A$ is the identity. It does what it is called: it stabilizes $A$ and $G$, in some sense of the word.

The set of stabilizing automorphisms might seem like a pretty restricted and elite class of automorphisms, but in fact just considering these will be enough to prove that $D_n$ has an outer automorphism when $n$ is even. That is, the outer automorphism of $D_n$ that we will show to exist is a stabilizing automorphism!

How can we see this, though? We use $H^1(G,A)$! Suppose we have a derivation $D:G\to A$. Recall that this just means that $D$ is a function and $D(gh) = D(g)h + D(h)$ for all $g,h\in G$. I assert that this derivation gives rise to a stabilizing automorphism of $A\rtimes G$! (Can you guess how we are going to show that $D_n$ has an outer automorphism?)

In fact, if $D$ is a derivation then the map $(a,g)\mapsto (a + D(g),g)$ is in fact a stabilizing automorphism. This is a short verification, after we observe that $D(1) = 0$. Now, the reader should also check the following facts, which are not difficult:

• The stabilizing homomorphisms correspond bijectively with the derivations
• The inner stabilizing homomorphisms correspond bijectively to the principal derivations

Hence outer stabilizing automorphisms arise from derivations that are not principal derivations. Thus $A\rtimes G$ having an outer automorphism is equivalent to $H^1(G,A)$ being nontrivial.

### Computation

So, we just need to show that $H^1(C_2,C_n)$ is nontrivial if $n$ is even in order to show that $D_n$ in this case has an outer automorphism. Luckily, computing cohomology and homology for cyclic groups is straightforward. Recall that $H^1(C_2,C_n)$ is just the first derived functor of $-^{C_2}$ applied to $C_n$. Does this mean we have to find an injective $\mathbb{Z}C_2$-resolution of $C_n$ in the category of $C_2$-modules? Finding injective resolutions is not always easy. Luckily for us, we can employ a nice little trick. For any group $G$, the functor $-^G$ is naturally equivalent to $\mathrm{Hom}_G(\mathbb{Z},-)$. This isn't surprising because half-exact additive functors behave essentially like $\mathrm{Hom}$ or $\otimes$ (but this is something for another day).

Moreover, $\mathrm{Hom}(-,-)$ is a balanced bifunctor, which is a highfalutin way of saying that for any $X$ and $Y$ the derived functors arising from $\mathrm{Hom}(X,-)$ applied to $Y$ and the derived functors arising from $\mathrm{Hom}(-,Y)$ applied to $X$ are isomorphic. This can be proved using a spectral sequence, for instance (hooray for spectral sequences!).

Anyways, what this all means is that instead of taking an injective resolution of $C_n$ we can choose a projective $\mathbb{Z}C_2$-resolution of the trivial module $\mathbb{Z}$ and apply $\mathrm{Hom}_{C_2}(-,C_n)$ and the first cohomology of this complex will actually be $H^1(C_2,C_n)$. If $\sigma$ is the generator of $C_2$ and $N = 1 + \sigma\in \mathbb{Z}C_2$ ($N$ is called the norm element) then we just happen to have a nice periodic projective resolution available:

$\cdots \mathbb{Z}C_2\xrightarrow{\sigma-1} \mathbb{Z}C_2\xrightarrow{N} \mathbb{Z}C_2\xrightarrow{\sigma-1} \mathbb{Z}C_2\xrightarrow{\epsilon}\mathbb{Z}\to 0$

Here the maps are mulitiplication and the map $\epsilon:\mathbb{Z}C_2\to\mathbb{Z}$ is the augmentation map: an element $x= \sum_g a_gg\in \mathbb{Z}C_2$ is mapped to $\epsilon(x) = \sum a_g$. This resolution works for any cyclic group and not just $C_2$ of course.

Now the cohomology group $H^1(C_2,C_n)=H^j(C_2,C_n)$ for $j$ odd is just the usual kernel mod image of

$\cdots\xrightarrow{\circ(\sigma-1)}\mathrm{Hom}_{C_2}(\mathbb{Z}C_2,C_n)\xrightarrow{\circ N}\cdots$

But $\mathbb{Z}C_2$ is a free $\mathbb{Z}C_2$ module, so this complex reduces to

$\cdots\xrightarrow{\sigma-1}C_n\xrightarrow{N}\cdots$

If $x\in C_n$ then $Nx = (1 + \sigma)x = x – x = 0$. On the other hand $(\sigma – 1)x = -2x$. Thus we get

$H^1(C_2,C_n) = C_n/2C_n.$

So if $n$ odd then $2C_n = C_n$ whereas, if $n$ is even then $C_n/2C_n\cong \mathbb{Z}/2$! Hooray! Thus $H^1(C_2,C_n) \cong \mathbb{Z}/2$ if $n$ is even, so in this case $C_n\rtimes C_2\cong D_n$ has an outer automorphism!

### Conclusion

I admit, the above exposition might seem at first to be a bit long. In fact, since $D_n$ is a finite group we could have found the outer automorphism by brute force, which of course would hardly be any fun at all.

However, I assure the reader that once she becomes familiar with the standard tools of homological algebra and group cohomology, the above thought process will become encapsulated in a few lines and a moment of thought; although this is indubitably a large piece of machinery to expose that poor outer automorphism of $D_n$ for $n$ even, the wide range of applications of group cohomology (and homological algebra!) to other topics of mathematics like group extensions, class field theory, modular representation theory, and algebraic topology will certainly make learning this topic worthwhile.