Groups Lecture 11

in which we explore groups of order 6 and make a start on quotients.

Last time we defined normal subgroups as those for which each left coset $aK$ is the same as the right coset $Ka$, proving that any subgroup of index $2$ is normal, as is every subgroup of an abelian group. Today we added that every kernel is normal. We will see soon that in fact kernels are exactly the normal subgroups (meaning every normal subgroup is a kernel of some group hom, and every kernel is a normal subgroup). As promised last chapter, we proved that every group of order $6$ is cyclic or dihedral. This uses Lagrange and normal subgroups.

Then we moved to quotients, which take most students a while to understand (so do persevere please). We saw that, for a normal subgroup $K \trianglelefteq G$, we can define a group operation $aK*bK=(ab)K$ on the set of left cosets. The main thing to show here was that this is well-defined, and that relies crucially on $K$ being a normal subgroup. The resulting group is called the quotient group (or factor group) and written $G/K$. We saw a few examples for $G=\mathbb{Z}$, $G=D_6$ and $D_8$. To reinforce the fact that we need a normal subgroup to be able to form quotients, we looked at the example of $\langle s\rangle \leq D_6$, which is not normal. The set of left cosets still exists, but if we try to multiply two elements using representatives, we get different answers for different choices, so the operation is not well-defined.

We defined the very important quotient map $q\colon {G\to G/K}$ which takes $a\in G$ to the coset $aK$. It is a surjective group homomorphism. This helped us to see that (subgroups and) quotients of cyclic groups are always cyclic. We ended the lecture with some warnings: Quotients are not subgroups of $G$!!!! They may not even be isomorphic, as this example shows: all non-trivial subgroups of $\mathbb{Z}$ are infinite (they are $n\mathbb{Z}$), and all proper quotients of $\mathbb{Z}$ are finite (they are $\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_n$). (The trivial ones here are $0\mathbb{Z}=\{0\}$ and $\mathbb{Z}/0\mathbb{Z}\cong \mathbb{Z}$.)

Understanding today’s lecture

Quotients need getting used to; many people still haven’t understood them in second year. But you will understand them better if you play around with them a lot. When dealing with quotients, remember that they are not subgroups! The elements are different; if $G$ has elements $a\in G$, then a quotient $G/K$ has cosets $aK$ as elements, which are themselves subsets of $G.$ So you have a set of subsets of $G$, rather than a set of elements of $G$. You can often calculate with them fairly easily through a representative, because after all the operation is defined as $aK*bK=(ab)K$, but you have to remember that every coset has several representatives you could have chosen. If you are more comfortable with the $[a]$ notation we had for equivalence classes, maybe that might help. If you want to remind yourself of the “motivational game” we played on the blackboard, you can look it up in the type-set lecture notes.

Preparing for Lecture 12

We will keep going with quotients and prove the Isomorphism Theorem, which relates the image of a homomorphism to the quotient of the domain by the kernel of the homomorphism. So make sure you get some practice with quotients! We will also finally prove what we’ve been alluding to for ages: the essential uniqueness of cyclic groups. So remind yourself of the definition of a cyclic group.

We will start our next chapter next time: Group Actions. You might want to make yourself a cube to bring for visualisation purposes. Here is a net of a cube with lots of useful things on it that I will be referring to a lot in the next few lectures: the Taylor-Cube (designed by Gareth Taylor).

Going a little deeper

We have seen that every kernel is a normal subgroup. And we have seen by now quite a few subgroups which are not normal. So what can we deduce from that? Those non-normal subgroups can never be a kernel of any group homomorphism! If we are working in abelian groups, all subgroups are normal, so we don’t have this problem. But in general, it is quite an important point. In my research area I work exactly in such situations, where not all injections are kernels of some homomorphism.

You can define kernels another way, with a universal property. I had a professor in Germany who made us all learn about universal properties very early on. They can be very useful! (Though you only learn about it very formally in fourth year here.) If you have a group homomorphism $f\colon {G\to H}$, then some group homs $g\colon { L\to G}$ have the property that $f(g(a))=e$ for all $a\in L$. We say that the composite $fg$ is the zero morphism (it sends everything to the identity, which often happens to be $0$, that is where the name comes from). The kernel $K$, viewed as an injection $K\to G$, is such a hom, but a very special one: every other $g$ with that property factors through the kernel, which means that for all $a\in L$ we have $g(a)\in K$. The image of $g$ is inside the kernel $K$ of $f$. So the kernel is a universal hom with this property. We will see next time that quotient groups also have such a universal property (but “the other way round”).