# Groups Lecture 12

in which we prove the Isomorphism Theorem and first meet group actions.

Today we proved the Isomorphism Theorem: if $f\colon {G\to H}$ is a group hom with kernel $K$, then $K\trianglelefteq G$ (as seen before) and $G/K\cong \mathrm{Im}f$. We proved this by defining a homomorphism $G/K\to H$ that sends any coset to the image (under $f$) of any of its elements. Again, as usually with cosets, the main thing is to check that this is well-defined. The reason behind it is that the cosets of the kernel give you exactly the partition of $G$ into the subsets which get mapped to the same image by $f$. Once we saw that our map was well-defined, we could prove that it is indeed an injective group hom, giving us an isomorphism to the image of $f$ as required. As a consequence, when we have an injection $f\colon {G\to H}$ (i.e. the kernel is just the identity element), we can view $G$ as (isomorphic to) a subgroup of $H$. And if $f$ is a surjection, then the codomain $H$ itself is a quotient of $G$. A possible slogan for the Isomorphism Theorem could be “homomorphic images are quotients“. As examples we looked at the determinant of an $n\times n$ matrix, we found the unit circle as the quotient $\mathbb{R}/\mathbb{Z}$, and had a quick look at how many “quadratic residues” there are modulo a prime $p$.

Then we used the Isomorphism Theorem to show essential uniqueness of cyclic groups: any cyclic group is isomorphic to either $(\mathbb{Z},+)$ or $(\mathbb{Z}/n\mathbb{Z},+_n)$, for some $n\in \mathbb{N}$.  We also defined a simple group as one which only has $\{e\}$ and itself as normal subgroups. Examples are $C_p$ for $p$ prime (by Lagrange; in fact they have no other subgroups at all, let alone normal ones), and $A_5$, the alternating group on $5$ elements, which we will investigate more in Chapter 6.

After that we started on Group Actions (Chapter 5). An action of a group $G$ on a set $X$ is a function $\theta\colon {G\times X\to X}$ which satisfies:

0. $\theta(g,x)\in X$ for all $g\in G, x\in X$;
1. $\theta(e,x)=x$ for all $x\in X$;
2. $\theta(g,\theta(h,x))=\theta(gh,x)$ for all $g,h\in G, x\in X$.

We saw several other alternative notations to write this $\theta$, for example $\theta(g,x)=g\cdot x=g(x)$, or we can view $\theta(g, - )\colon {X\to X}$, for a fixed $g\in G$, as a map $\theta_g\colon {X \to X}$. We will prove next time that these $\theta_g$ are all bijections. As examples for actions we saw the trivial action, which does nothing, $S_n$ and $D_{2n}$ both acting on $\{1,2,\ldots, n\}$, and rotations of a cube acting on several different sets, such as the faces of the cube, the diagonals, the “pairs of opposite faces”, etc.

Understanding today’s lecture

The Isomorphism Theorem is one that we will use over and over again, and eventually we will probably not even notice any more when we are using it. It is that central. You can get used to it by applying it to any group hom you have lying around at home (or meet on your travels).

Actions are what make groups tangible or concrete. Through actions, the groups are really “doing” something. We have had several different notations: see that you can translate between them. And maybe you have a favourite one? In lectures we will use different ones in different contexts, as they all have their advantages and disadvantages, and it depends what you want to do as to which is most useful.

We will talk more about rotations of a cube as well in future lectures. I suggest you make yourself your own “Taylor-cube” to play around with, you can find the net to print out here. You may very well want to check out Gareth Taylor’s website which has many more useful things on it.

Preparing for Lecture 13

Next time we will keep exploring actions, by giving an alternative definition and finding out a bit more about them. In particular, we will be looking at orbits and stabilisers. Orbits are all elements of the set that can be reached from a given element, and stabilisers are all group elements which fix a particular element. You can read up on these in any group theory book (or the latexed notes) before next time. We will prove that orbits partition the set $X$, and the very important Orbit-Stabiliser Theorem, which is extremely useful for finding sizes of all sorts of things. The proof is a sort of “geometric version of Lagrange”, so go and freshen up on the proof of Lagrange.

Going a little deeper

The Isomorphism Theorem isn’t really something that is restricted to groups, there is a version of it in many other areas of maths. (And in fact there are other Isomorphism Theorems for groups as well, sometimes called Noether’s Isomorphism Theorems. But our one is “the” (or first) Isomorphism Theorem.) In second year Groups Rings and Modules you’ll see one for rings and one for modules. Either in Vectors and Matrices, or at least in second year Linear Algebra, you will see one for vectorspaces, though it looks different (and a lot easier, because vectorspaces are very nice). The rank-nullity theorem says: if $\alpha\colon {V\to W}$ is a linear map between vectorspaces, then $\mathrm{dim}V=\mathrm{rank}(\alpha)+\mathrm{nullity}(\alpha)$, where the rank is the dimension of the image and the nullity is the dimension of the kernel. If you compare that to our isomorphism theorem and (once you know more about finite dimensional vectorspaces) see what the equivalent would be for vectorspaces, you’ll see it is exactly this.

So, in what sense are quotient groups “the opposite of” kernels? I explained last blog that kernels have a certain universal property. Quotients also have a universal property, but sort of on the other side of the homomorphism. Given $f\colon {G\to H}$ with kernel $\iota\colon {K\to G}$ (viewed as an inclusion), we know that $f\iota(k)=e$ for all $k\in K$. Writing $q\colon {G\to G/K}$ for the quotient map as in lectures, this $q$ has the same property: $q\iota(k)=e$ for all $k\in K$. Remember that we can describe this as “the composite $q\iota$ is zero”. Now if we have any other group hom $g\colon {H\to L}$ which also satisfies $g\iota(k)=e$ for all $k\in K$, then this $g$ factors through the quotient map, in the sense that there is a unique group hom $\overline{g}\colon {G/K\to L}$ satisfying $\overline{g}q=g$. We saw in the proof of the Isomorphism Theorem one special case of how such a morphism $\overline{g}$ might be defined. Perhaps you can extend that construction to this case? We won’t necessarily get that $\overline{g}$ is injective, like we had in the Isomorphism Theorem, but the rest works just the same.