# Groups Lecture 4

Today we proved some important properties of group homomorphisms: Group homomorphisms send identity to identity and inverses to inverses; the composite of two group homs is a group hom, and both the composite of two isomorphisms and the inverse of an isomorphism are again isomorphisms. We then defined two groups related to a group homomorphism $f\colon {G\to H}$: the image of $f$ is the set $\mathrm{Im}f=\{f(a) \mid a\in G\}$ of elements in $H$ “hit by” $f$, and the kernel of $f$ is the set $\mathrm{Ker}f=\{a\in G\mid f(a)=e_H\}$ of elements in $G$ that are mapped to the identity. We proved that the image is a subgroup of $H$ and the kernel is a (normal) subgroup of $G$. After investigating some examples, we also proved  how to show injectivity via kernels when we are working with group homomorphisms.

At the very end we met cyclic groups, which are groups generated by a single element, meaning that all elements of the group can be written as powers of the generator. The intergers are “the infinite cyclic group”, and $\mathbb{Z}_n$, integers mod $n$, are also cyclic. We will get back to them a bit more next time.

Today I handed out Example Sheet 1. You should now be able to do up to question 7, and if you use the definition of order given on the sheet, probably even up to 10. We will define the order of an element properly in lectures next time.

Understanding today’s lecture

Make sure you prove that group homomorphisms send inverses to inverses (that is on Practice Sheet C from last time). Play around with the group homomorphisms you have met (for example on Sheet C) and find their images and kernels. How might the “injectivity via kernels” result help you when dealing with group homomorphisms? Then have a look at generators of cyclic groups: can you find some “rule” which tells you what elements of $\mathbb{Z}_n$ are generators, starting from the examples given in lectures?

Going a little deeper

When we proved “the inverse of an iso is an iso”, what we were really proving was that if you have a bijective group homomorphism, the inverse function is also a group homomorphism. The “morally correct” definition of an isomorphism is really a group homomorphism $f\colon {G\to H}$ with an inverse group homomorphism $f^{-1}\colon {H \to G}$. But as we have seen, being a bijective group hom is enough to make this happen: a bijective group homomorphism has an inverse function, and this function is automatically a group homomorphism. This does not happen in all areas of mathematics. For example, in Analysis I in Lent term  you will learn about continuous functions. A bijective continuous function always has an inverse function, but the inverse need not be continuous! In Analysis you then call the ones that do have a continuous inverse homeomorphisms (the extra e there being the important bit), but one could also just call it “an isomorphism of topological spaces”. But you’ll only learn about that in Metric and Topological Spaces in Easter (or in second year). 🙂

Looking back to a comment from last time, we have seen from the properties of group homomorphisms that indeed they don’t only preserve the group operation, but also the identity and inverses. We had a similar comment with subgroups after Lecture 2. So what if we get rid of inverses again? Can you work out what properties a monoid homomorphism should have? Is it enough to ask $f(a*b)=f(a)\star f(b)$ for all $a,b\in M$ where $M$ is a monoid, i.e. it has a “multiplication” which is associative, closed and has an identity? Does such an $f$ automatically preserve the identy? Or do we have to ask that as an extra condition?

Preparing for Lecture 5:

Next time we will revisit cyclic groups (from today) and dihedral groups, the groups of symmetries of a regular $n$-gon. So remind yourself about these symmetries and what properties they have. Think about what happens if I give you a single element of some larger group and ask you to give me the smallest subgroup containing that element. What would it be? What if I gave you more than one element?