# Groups Lecture 5

in which we investigate cyclic groups, generators, dihedral groups and cartesian products of groups.

At the end of the last lecture, we defined cyclic groups, which are groups generated by a single element, so they consist only of powers (both positive and negative) of that element. Examples of cyclic groups are $\mathbb{Z}$, the “infinite cyclic group”, $\mathbb{Z}_n$, and $(\{1,-1\}, \cdot )$, and the rotations of a regular $n$-gon. We saw a relationship between the order of a group element and the cyclic subgroup generated by that element: the order of $a\in G$ is the smallest positive integer $k$ satisfying $a^k=e$, and it turns out to be the same as the size of the subgroup $\langle a \rangle$ generated by $a$ (so it could also be infinity). I left it to you to prove that all cyclic groups are abelian, and also that any subgroup of a cyclic group is cyclic. We then looked at the dihedral groups, symmetries of regular $n$-gons, in terms of generators and relations: $D_{2n}=\langle r, s\mid r^n=e=s^2, srs^{-1}=r^{-1}\rangle$. This gives us both a geometric viewpoint (as rotations and reflections of an $n$-gon), and an algebraic viewpoint (as elements that can be written as $r^k$ or $sr^k$ for some integer $k$). The geometric viewpoint is very useful to train our intuition and to work out what we expect should happen, but it is often easier to give a completely water-tight proof with algebra rather than with geometry. Combining both as appropriate gives us the best of both worlds. By the way, someone asked me after lectures whether we can have dihedral groups for $n=1$ or $2$. I think usually you would just start at $3$, because that is the first proper $n$-gon which is really a two-dimensional shape. For $n=1$ and $2$ the sizes don’t work out any more to be $2n$, so we usually leave those out.

We also looked at cartesian products of groups, that is, at the componentwise group operation $(a_1,a_2)*(b_1, b_2)=(a_1*_1b_1, a_2*_2b_2)$ on the set $G_1\times G_2=\{(a_1,a_2)\mid a_i\in G_i\}$ of ordered pairs. The most important thing we noticed about this is that “everything in $G_1$ commutes with everything in $G_2$“, because $(a_1,e)*(e,a_2)=(a_1,a_2)=(e,a_2)*(a_1,e)$. So we can never say that for example $D_6$ is isomorphic to $C_3\times C_2$, because the product is abelian and the dihedral group is not!

Understanding today’s lecture

You don’t have the luxury of a practice sheet from me any more, so now you have to find a way how you can get used to the lecture material yourself. Make sure you do the exercises given in lectures, such as showing that cyclic groups are abelian. Think about the order of generators: if you have a cyclic group with more than one generator, how are the orders of these possible generators related? For example you could take $\mathbb{Z}_n$ for a few small $n$ and write out the orders of all elements, and see which ones are generators and which ones are not, and why they are/are not.

Play around with the algebraic “generators and relations” representation of the dihedral group so you  get comfortable with those $r$s and $s$es, how to manipulate them given the relations, and so on. You could redo that question on Sheet B asking “Show that the symmetries of a regular triangle form a subgroup of the symmetries of a regular hexagon”. It will be easier to prove with $r$ and $s$! (I think so anyway.) Also play around with some cartesian products and see if you can find another group such a product is isomorphic to. We had $C_2\times C_2$ and $C_2\times C_3\cong C_6$. Especially think about the dihedral groups and what their relationships are with cyclic subgrous; we saw that they are not a cartesian product of cyclic groups!

Going a little deeper

Think about the connection between what we said about the order of an element being the size of the cyclic subgroup generated by that element, and the proof that $n\mathbb{Z}$ are the only subgroups of the integers $\mathbb{Z}$. Can you see a connection with this “smallest natural number” argument? Related is the exercise given in lectures that every subgroup of a cyclic group must be cyclic.

We said for the cartesian product $G_1\times G_2$ that “everything in $G_1$ commutes with everything in $G_2$“. This would only really make sense if they are subgroups of the cartesian product. They are not straight-forward subgroups, because we have pairs of elements now, but they are in fact isomorphic to subgroups in a very natural way: Taking the set $\{(a_1,e)\mid a_1\in G_1\}\subset G_1\times G_2$ gives us a subgroup which “is” (meaning is isomorphic to) $G_1$, and similarly $\{(e,a_2)\mid a_2\in G_2\}\subset G_1\times G_2$ is isomorphic to $G_2$. So in that sense $(a_1,e)*(e,a_2)=(e,a_2)*(a_1,e)$ can be read as “everything in $G_1$ commutes with everything in $G_2$.”

If we extend this to the dihedral groups, we see that $D_{2n}$ does have a subgroup which is isomorphic to $C_n$, namely the subgroup of rotations, but the rotations do not commute with all the reflections! So it is a different scenario and we can’t say $D_{2n}\cong C_n\times C_2$. For some $n$, however, there is one particular element of order two which does commute with everything in the group. Can you find it?

Preparing for Lecture 6:

Next time we will work a lot on the Symmetric Group. This is a very important example of groups, so we will spend quite some time on it and come back to it later.  To prepare the ground, so to speak, you could think about that question with “composition of functions is always associative” again. What special functions do we have to restrict to to get a hope of having a group with composition as group operation? For example, can you say something about what the domain and codomain could be? Can you say anything more about which sort of functions we need to look at?

If you feel so inclined, you might also look up cycle notation in a groups book, for example Beardon Algebra and Geometry (as given in the schedules). That is usually easier the second time you see it :-).