in which we find out more about quotients, including the quotient map, and prove the Isomorphism Theorem.
We are continuing our investigation of quotients. Today we saw that the group operation on cosets only works when we have a normal subgroup, by looking at the example of a reflection in . We met the very important quotient map which exists for any normal subgroup . This helped us to see that (subgroups and) quotients of cyclic groups are always cyclic. We then stated the Isomorphism Theorem: if is a group hom with kernel , then (as seen before) and . We proved this by defining a homomorphism that sends any coset to the image (under ) of any of its elements. Again, as usually with cosets, the main thing is to check that this is well-defined. And once we saw that, we could prove that it is indeed injective, giving us an isomorphism to the image of as required. As a consequence, when we have an injection (i.e. the kernel is just the identity element), we can view as (isomorphic to) a subgroup of . And if is a surjection, then the codomain itself is a quotient of . A possible slogan for the Isomorphism Theorem could be “homomorphic images are quotients“. As examples we looked at the determinant of an matrix, we found the unit circle as the quotient , and had a quick look at how many “quadratic residues” there are modulo a prime .
Understanding today’s lecture
When dealing with quotients, remember that they are not subgroups! The elements are different; if has elements , then a quotient has cosets as elements, which are themselves subsets of . You can often calculate with them fairly easily through a representative, because after all the operation is defined as , but you have to remember that every coset has several representatives you could have chosen. And it is always good to keep this example in mind: all non-trivial subgroups of are infinite (they are ), and all proper quotients of are finite (they are ). (The trivial ones here are and .)
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.
Preparing for Lecture 13
Next time we will look at cyclic groups again, proving that they are essentially unique. That means, up to isomorphism, there is only one cyclic group of any given order. It seems intuitive, but the Isomorphism Theorem and some other results we’ve had will help us prove it very rigorously. So it would be good if you can remind yourself about everything you know about cyclic groups and about subgroups of before next time.
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 is a linear map between vectorspaces, then , 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 with kernel (viewed as an inclusion), we know that for all . Writing for the quotient map as in lectures, this has the same property: for all . Remember that we can describe this as “the composite is zero”. Now if we have any other group hom which also satisfies for all , then this factors through the quotient map, in the sense that there is a unique group hom satisfying . We saw in the proof of the Isomorphism Theorem one special case of how such a morphism might be defined. Perhaps you can extend that construction to this case? We won’t necessarily get that is injective, like we had in the Isomorphism Theorem, but the rest works just the same.