# 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. If you want to know about some “strange cases” of dihedral groups for $n=1$ or $n=2$, look in “Going deeper” below.

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!

We have now covered all the material you need for the first example sheet.

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.) Showing that all elements of the form $sr^k$ have order $2$ is also a very good exercise to start getting used to manipulating these elements. 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!

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 :-).

Going a little deeper

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 $n\geq 3$). For some $n$, however, there is one particular element of order two which does commute with everything in the group. Can you find it?

Our definition of dihedral groups as symmetries of regular $n$-gons implies that we might only want to start with $n=3$ (what is a $1$-gon or a $2$-gon after all?). But the definition given this time with generators and relations can make us think about the cases $n=1$ and $n=2$ as well. What group do we get if we just take the definition and put those values of $n$ in? For $n=1$ we get: $D_2=\langle r,s \mid r^1=e=s^2, srs^{-1}=r^{-1}\rangle$. The $r^1=e$ means in fact $r=e$ and we can just disregard it, so what we’re left with is $D_2=\langle s\mid s^2=e, s s^{-1}=e\rangle= C_2$. For $n=2$ we get: $D_4=\langle r,s \mid r^2=e=s^2, srs^{-1}=r^{-1}\rangle$. Now $r^2=e$ means that $r=r^{-1}$, so the equation $srs^{-1}=r^{-1}$ really becomes $sr=rs$. So we get the abelian group $D_4\cong C_2\times C_2$.

If you want some intuition about how to think of these as “symmetries of $n$-gons”, you have to indulge me in a bit more detail: Think of a single point but with an “orientation”, so you could imagine a little circle round it with an arrow pointing anti-clockwise. Then “reflecting” the point will still give you a point, but the little arrow now points clockwise. In this sense you could view $C_2$ as a symmetry of an oriented point (which we for this purpose could view as a $1$-gon). If you now imagine a line with two endpoints $1$ and $2$, each also with a little orientation arrow, you can imagine rotating the line around its midpoint to get another line with the end points swapped but the orientations the same, and reflecting the line in the perpendicular through the midpoint gives you a line with the end points swapped but the orientations opposite. Then if you do both after another (in either order), you’d get the line with the endpoints the same but the orientations opposite (that is like having reflected the line in itself). So in that way you can view $C_2\times C_2$ as the symmetries of a line with oriented endpoints (a $2$-gon). Another way to think of it for the line would be to say the line has a “top” and a “bottom”, and then rotation switches the endpoints and “top” and “bottom”, reflection in the perpendicular switches the endpoints but leaves “top” and “bottom” the same, and reflection in the line itself leaves the endpoints the same but switches “top” and “bottom”. But I like the orientation of endpoints version better.

And if you think about it a bit, the “orientation” of points comes naturally on any higher $n$-gon, we just normally see it through the orientation of the numbering on the whole $n$-gon: rotations don’t change the orientation, whereas reflections do. In the strange low cases we just have the unusual situation of having the numbering the same but the orientation changed. However, in most results about dihedral groups, we assume that $\geq 3$, because these cases are a bit weird: they are abelian for example, whereas all higher dihedral groups are not.