Presentation(s)!

The 11th group theory lecture

Last time, we introduced the notion of a normal group and how it can be used to define quotients.

The Importance of Normality

We also saw how if you have a group homomorphism φ: G→H, then the kernel of φ is a normal subgroup. But what is G/ker(φ)? We didn’t say, nor did we look at any examples. Let’s fix that.

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Theorem: (The First Isomorphism Theorem) Let φ: G→H be a group homomorphism. Then

\(\begin{align*} \tilde{\varphi}:G/\ker(\varphi) &\rightarrow \text{im}(\varphi) \\ g\ker(\varphi) &\mapsto \varphi(g) \end{align*}\)

is a well-defined group isomorphism.

Proof: Why is this well-defined? After all, if we choose g and h such that g ker(φ) = h ker(φ), must it be the case that φ(g)=φ(h)? Yes, because we know that gh^-1 is in the kernel, hence φ(gh^-1)=1, from which we deduce φ(g)=φ(h).

Is it a group homomorphism? Yes, it inherits that from φ. Is it surjective? Yes, because we are mapping into the image, which is by definition everywhere where φ sends something. Is it injective? Well, if g ker(φ) is in the kernel, then that means that φ(g)=1, which is to say that g is in ker(φ). □

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You might remember that way back when we introduced group homomorphisms, I noted that group theory has an analog of the rank-nullity theorem in linear algebra, and that we will have a means of turning arbitrary homomorphisms into isomorphisms via quotienting.

Homomorphisms and the Guises of Groups

We have arrived at that moment! Let’s look at some examples.

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Definition: Let G, H be groups. The product group G×H is the set of pairs (g, h) where g is in G and h is in H, with the product (a, b)(c, d) = (ac, bd).

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It is easy to check that G×H is also a group. It is also immediate that we have a surjective group homomorphism

\(\begin{align*} \pi:G\times H &\rightarrow G \\ (g,h) &\mapsto g. \end{align*}\)

What is the kernel? It is precisely the set of all pairs (1, h). Ergo, G×H/ker(π)≅G, which is not altogether surprising.

Here is a more interesting example. Let’s say that G is generated by elements in X. Let F_X be the free group generated by elements in X. Consider the map

\(\begin{align*} \phi: F_X &\rightarrow G \\ x_1^{k_1}\ldots x_n^{k_n} &\mapsto x_1^{k_1}\ldots x_n^{k_n}. \end{align*}\)

Observe: this is a group homomorphism! Furthermore, it is surjective, since we know that G is generated by words in X. What is its kernel? Well, it consists of all the reduced words in X that collapse to 1 in G.

We have thus demonstrated that every group G is isomorphic to a quotient of a free group, and we even have an idea for how to get this quotient—you need to think about all the relations in the generators. (E.g., if you know that x₁x₂=x₃x₁, you need to add the word x₁x₂x₁^-1x₃^-1 to the normal subgroup.)

This gives a very powerful way to construct groups from scratch, but we need a definition and a simple lemma to make full use of it.

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Definition: Let X be a subset of a group G. The normal closure ncl_G(X) of X in G is

\(\bigcap_{X \subset N \triangleleft G} N,\)

where N◁G means that N is a normal subgroup of G.

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Lemma: ncl_G(X) is a normal subgroup of X. (And is, therefore, the smallest normal subgroup that contains X.)

Proof: I leave this to the reader. □

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Definition: Let X be any set—we call these the generators. Let R be any set of words in X inside the free group F_X—we call these the relations. We shall write

\(\langle X | R \rangle := F_X/\text{ncl}_{F_X}(R).\)

If a group G is isomorphic to ⟨X|R⟩, then we say that it is a presentation of G. If one can find finite sets X and R such that G≅⟨X|R⟩, then we say that G is finitely-presentable.

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Observe that we have already proved that any group has a presentation: you just take X to be the set of generators for the group (we could take the entire group here), and you take R to be the words in those generators that are the identity in that group. But we can often get away with much shorter presentations.

While we define presentations in terms of normal closures, that is usually not the best way to think about it intuitively: a presentation is what we get if we take all words in the given set and enforce the relations—that is, the relations tell us what words should be equal in our group.

Let’s look at some examples.

1. ⟨X⟩ (that is, R is just the empty set) is simply F_X, because we don’t actually quotient out by anything.¹ So one way to think about a free group is that it is what we get by enforcing no relations.

2. ⟨x|x^N⟩ is just ℤ/Nℤ in disguise: after all, it is the set of words x^m, but where we force x^N=1.

3. ⟨x,y|xyx^-1y^-1⟩ is just ℤ². Why? The relation just tells us that x and y must commute. Therefore, this can be understood as the set of words x^myⁿ, where m and n are integers. If we multiply two such words x^myⁿ, x^ay^b, the result is x^m+ay^n+b. Thus, there is an obvious isomorphism (m,n)↦x^myⁿ.

4. The standard presentation of D₈ is ⟨R,T|R⁴, T², (TR)²⟩.

We should inspect this last example more closely. Certainly, we know that R⁴=T²=(TR)²=1 in D₈—we can see it from the geometry. (The first is a rotation, and the second and third are reflections.) Therefore, certainly we can say that

\(D_8\cong \left\langle R,T\middle|R^4, T^2, (TR)^2,\ldots\right\rangle,\)

but we cannot yet be certain that there are no additional relations that are required. For instance, we know that TR=R³T—do we know that this can be deduced from the given relations? Yes, because we know that (TR)²=1, which you can rearrange into TR=R³T once you observe that T^-1=T and R³=R^-1. (These follow from T²=R⁴=1.) Similarly, we can deduce RT=TR³ (just multiply by R on the left and R³ on the right). But maybe we are still missing some other relations?

Let’s observe the following: from what we have said, D₈ can be constructed by starting with ⟨R,T|R⁴, T², (TR)²⟩, and then possibly quotienting by some additional relations. Therefore, we have a surjective group homomorphism

\(\begin{align*} \left\langle R,T \middle| R^4, T^2, (TR)^2\right\rangle &\rightarrow D_8 \\ R&\mapsto R \\ T&\mapsto T. \end{align*}\)

We just need to show that this is an isomorphism. One way to do it is to observe that D₈ has 8 elements in it—therefore, if the group on the left has 8 elements, then this must be an isomorphism. (If it isn’t injective, we won’t get the full 8 elements in the image.)

What elements should we expect? Certainly, 1, R, R², R³, T, TR, TR², TR³—that is, the elements of D₈. Since they are distinct in the quotient (D₈), they must be distinct in the original group. But can everything else be reduced to one of those elements?

Observe that since T²=R⁴=1, we may assume that all powers of T are either 0 or 1, and all powers of R are 0, 1, 2, or 3. Furthermore, since RT=TR³, if we are given an arbitrary word in T and R, we can use this relation to iteratively push all T’s to the left, and all R’s to the right. E.g.,

\(\begin{align*} TR^2 T R^2 TR&=TRTR^3 R^2 TR \\ &=TRTRTR \\ &=TTR^4TR=TR. \end{align*}\)

Therefore, any word in R and T can be put in the form T^kR^m where T=0,1 and m=0,1,2,3… but those are exactly the elements of D₈! So, we are done.

Let’s build a new group entirely from scratch. Consider the quaternion group

\(\begin{align*} Q_8:=&\left\langle-1,i,j,k\middle|(-1)^2,(-1)i^2,\right. \\ &\quad \left.(-1)j^2,(-1)k^2,(-1)ijk\right\rangle. \end{align*}\)

Now, to avoid any confusion: a priori, -1 is just another symbol (like x), and the operation on this group is multiplication and not addition (we’ll see in a second that it isn’t abelian). With that said, one way to construct this group is to start with the quaternions and take this subset—in that case, -1 should indeed be understood as the additive inverse of 1. We’ll use the shorthand of writing -x to mean (-1)x, but first it is a good idea to prove that -1 commutes with everything (and so, in particular, this is the same as x(-1)).

To prove that -1 commutes with all elements in Q₈, it is sufficient to show that it commutes with all of the other generators. But (-1)i² can be rearranged into i²=-1 (using the fact that (-1)^-1=-1, which follows from (-1)²=1), which means that -1 is an element of ⟨i⟩, which we know is abelian—therefore, -1 commutes with i. The same argument—mutatis mutandis—applies to j and k.

That is useful, but we still need to understand what is happening with i, j, and k. We might observe that i⁴=j⁴=k⁴=(-1)²=1, so we should never consider any exponents other than 0, 1, 2, and 3. In fact, since i²=-1 and i³=-i (and similarly for j and k), there is actually no point in considering any exponents other than 0 and 1. So, all of our words are just strings of i’s, j’s, and k’s, with no higher exponents, and possibly with a -1 at the front.

Just as with D₈, we need to understand how (and if!) we can permute i, j, and k in such words. The needed rule comes from the fact that ijk=-1, because if we multiply by -k=k^-1 on the right, this becomes ij=k. This means, immediately, that we can eliminate k entirely—we can rewrite it in terms of i and j!

But how can we switch i and j? Well, (ijk)^-1=k^-1j^-1i^-1=-kji=-1, so if we multiply by k on the left, we get ji=-k. Ah ha! So we can freely switch i and j, as long as we pick up an extra minus sign. Therefore, at most, Q₈ consists of the elements ±1, ±i, ±j, ±k=±ij. However, how can we know that all of these 8 elements are different? Maybe some of them are the same, and we just haven’t been clever enough with our relations?

That is not the case, and we can prove it using a representation.²

Here’s how: let’s build a group homomorphism Q₈→GL(2, ℂ) by specifying that we send

\(\begin{align*} -1 &\mapsto \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \\ i &\mapsto \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \\ j &\mapsto\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \\ k &\mapsto \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}. \end{align*}\)

(These are the rescaled Pauli matrices, which are connected to the measurement of spin in quantum mechanics.)

Is this well-defined? Well, certainly we can define a group homomorphism F_{-1,i,j,k}→GL(2, ℂ) this way.³ The question is whether adding the relations for Q₈ will break things. But there is an easy way to check: just see whether those same relations are satisfied in GL(2, ℂ)! And they are:

\(\begin{align*} (-1)^2 = 1 &\rightsquigarrow (-I)^2 = I \\ i^2 = -1 &\rightsquigarrow \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}^2=-I \\ j^2 = -1 &\rightsquigarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^2=-I \\ k^2 = -1 &\rightsquigarrow \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}^2=-I \\ ijk=-1 &\rightsquigarrow \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \\ &\quad\quad=-I \end{align*}\)

So, there is no problem defining a group homomorphism Q₈→GL(2, ℂ) this way—everything that is identified on the left-hand side will be identified on the right-hand side. However, we can see that the images of ±1, ±i, ±j, ±k=±ij are all different matrices. Therefore, the original elements must have been distinct!

We conclude that Q₈={±1, ±i, ±j, ±k=±ij}.