# Groups Lecture 18

in which we meet the quaternions, start exploring matrix groups and act with matrices on vectors and on matrices.

We started last lecture with the result on groups of order $8$ and finished the proof today: Any group of order $8$ is either abelian and iso to one of $C_8, C_4\times C_2, (C_2\times C_2)\times C_2$, or non-abelian and iso to one of  $D_8$ (dihedral) or $Q_8$ (quaternion). We proved this by looking at elements of certain orders: if we have an element of order $8$ or only elements of order ($1$ and) $2$, it is easy. Given an element $a$ of order $4$ and an element $b$ which is not in $\langle a\rangle$, we used the fact that $\langle a\rangle$ is normal to show that $b^2$ (commutes with $a$ and) is either $e$ or $a^2$. Adding the restriction $bab^{-1}=a$ or $a^{-1}$ gives us four cases, which we investigated.

One of those four cases gave us some new relations we didn’t know about yet, which might lead to a new group (or to some contradictions). We saw by looking at a certain matrix group that such a group, called quaternions, does indeed exist, and we can write it out in several different ways.

Having mentioned them a few times in the beginning of the course, we finally properly defined the general linear groups $\mathrm{GL}_n(F)$ (for $F=\mathbb{R}$ or $\mathbb{C}$) as those matrices over $F$ which are invertible. The determinant gives us a group homomorphism which has kernel $\mathrm{SL}_n(F)$, the special linear group. I left you to work out the quotient $\mathrm{GL}_n(F)/ \mathrm{SL}_n(F)$ using the isomorphism theorem.

We then looked at two different actions of $\mathrm{GL}_n(\mathbb{C})$ ($\mathrm{GL}_n(\mathbb{R})$ works similarly). The group $\mathrm{GL}_n(\mathbb{C})$ acts on $\mathbb{C}^n$ faithfully with two orbits, by viewing the matrices as linear maps, i.e. “applying the function to a vector”.  As they are all invertible linear maps, $0$ forms a singleton orbit and all the other vectors form an orbit together. You were challenged to find out what orbits there are under the action of $\mathrm{SL}_n(\mathbb{C})$ on $\mathbb{C}^n$.

Another (possibly more interesting) way that $\mathrm{GL}_n(\mathbb{C})$ acts is by conjugation on the set $M_{n\times n}(\mathbb{C})$ of $n\times n$ matrices. We can think of this as base change: two conjugate matrices represent the same linear map $\mathbb{C}^n\to \mathbb{C}^n$ with respect to different bases. We will look at this again next time and see the types of orbits that this action has.

Understanding today’s lecture

The “groups of order 8” proof is quite long and convoluted with lots of cases, so it needs some time to get your head round it. Have a look at what techniques we used, and whether you’ve seen them/used them in other contexts. We had the Direct Product Theorem several times, we used normality, and we used conjugation quite a lot to find out what possible relations between $a$ and $b$ there can be. Perhaps you can write out a short summary, or bullet points of the important steps, to help you get an overview of the proof.

The material in the matrix groups chapter has a lot of intersection with and relevance for Vectors and Matrices. You will benefit from looking at both sources together, seeing what results from one course you use/need in the other, how they might give different viewpoints or proofs for the same thing, how they build up a better picture for you of understanding matrices as linear maps.

Preparing for Lecture 19

Next lecture we will look at orthogonal groups and prove that orthogonal matrices are isometries. So brush up on the transpose of a matrix and get your “geometry in two and three dimensions” cap on.

Going a little deeper

In the proof of the “groups of order 8” result, we boldly claimed that if we have $a$ of order $4$ and $b$ of order $2$ with $bab^{-1}=a^{-1}$, then our group is (isomorphic to) $D_8$. How can we be so sure? What if there are some other relations lurking that we just haven’t found out about? In general, given a group in terms of generators and relations, deciding whether two specific products of a sequence of generators (also called “words”) are the same group element or not can be undecidable. (Look up “word problem” for groups to find out more, but it is quite advanced.) But we are in a case which is manageable. The $bab^{-1}=a^{-1}$ relation, equivalently written as $ba=a^{-1}b$, tells us how to “commute $b$s past $a$s”, so we can put all the $b$s at the front, where they will cancel each other out to leave either no or one $b$. The $a$s at the end also reduce to $a^k$ for $k\in \{0,1,2,3\}$. So we can easily check that $\langle a,b\mid a^4=e=b^2, ba=a^{-1}b\rangle$ has exactly $8$ elements, and our group also has $8$ elements, so we must be ok. More formally, what we are doing is defining a group homomorphism from $D_8=\langle r,s\mid r^4=e=s^2, sr=r^{-1}s\rangle$ to our group of order $8$, sending $r$ to $a$ and $s$ to $b$. The point of all those relations is to check that this really does give a group homomorphism (cf properties of group homs and the related example sheet questions). Then we see it is surjective onto our group of size $8$, so as we have finite groups of the same size, it must be a bijection, i.e. also injective.