# Groups Lecture 11

in which we define normal subgroups, explore groups of order 6 and make a start on quotients.

We saw last time that not all subgroups behave the same with respect to left and right cosets. Today we defined normal subgroups as those for which each left coset $aK$ is the same as the right coset $Ka$. We can also say: for any $a\in G$ and $k\in K$, we have $aka^{-1}\in K$. We will see later that this could be called “stable under conjugation” or something like that. We saw that any subgroup of index $2$ is normal, as is every subgroup of an abelian group, and every kernel. 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. (I’ll write an alternative proof in a comment, so people who want to find it themselves don’t have to look before they’ve tried it.) The resulting group is called the quotient group (or factor group) and written $G/K$. We saw a few examples for $G=\mathbb{Z}$ and $G=D_6$.

Understanding today’s lecture

Normal subgroups are very important. Make sure you are happy with the equivalence of the three different ways of “defining” or checking whether something is a normal subgroup. You will probably find different ones useful in different contexts. Go through all the subgroups you’ve met so far (for example in the earlier lectures, or on the practice sheets, and on the example sheets) and find out which ones are normal.

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. Elements of quotients are cosets, that can be hard to understand, but practice will make it seem easier. If you are more comfortable with the $[a]$ notation we had for equivalence classes, maybe that might help.

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!

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 $f(g(a))=0$ for all $a\in L$. We say that the composite $fg$ is the zero morphism (it sends everything to zero). 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”).

If $a_1K=a_2K$ and $b_1K=b_2K$, then by the “same coset check” we know that $a_2^{-1}a_1\in K$ and $b_2^{-1}b_1\in K$. What we want to get to is $(a_2b_2)^{-1}(a_1b_1)=b_2^{-1}a_2^{-1}a_1b_1\in K$. So lets see if it is. $b_2^{-1}a_2^{-1}a_1b_1=b_2^{-1}kb_1$ because $a_2^{-1}a_1=k\in K$, and then $b_2^{-1}kb_1=b_2^{-1}b_1 b_1^{-1}kb_1=b_2^{-1}b_1 k'$ for some $k'\in K$ because $K$ is a normal subgroup. And since also $b_2^{-1}b_1\in K$, the whole prodcuct is in $K$, so we get $a_1b_1K=a_2b_2K$ as desired.