The aim of this post is to familiarise you well enough with Cayley tables so that in the next post we can really pin down what it means for two groups to be the ‘same’.

**A brief aside: a recap of information we can find from a Cayley table.** There are two key things to note. First, how would you compare gh and hg for all pairs g, h in a group G? If these were always equal (i.e. gh=hg for all pairs) then the group, by definition, would be commutative: all pairs would commute. This can be seen in the Cayley tables below: both of the them can be seen to be commutative by the fact that they are symmetric through the diagonal running top left to bottom right. The order of an element can be computed by running through each power e.g. in the left table we can compute the powers to see that (90°)^{1}=90°, (90°)^{2}=90°∗90°=180°, (90°)^{3}=180°∗90°=270°, (90°)^{4}=270°∗90°=0°, meaning that 90° is of order 4.

**Plenty of examples.** Previously, we have that the integers modulo m with addition produce the same group as C_{m}. We can see this by drawing the Cayley tables. Remember red then blue, not that this matters for these commutative groups. We can then say that these have the same structure since the difference between the group tables is just a change of notation. You might say: “they’re the same. You just divide the angles by 90 and you get the Cayley table for the integers modulo 4”. What about the next two tables? It might be worth deciding if it is a group by finding the identity and inverses. Is it commutative? Is it cyclic? Have a go before reading on.Fair enough, that one was a bit harder. The left one is isomorphic to C_{4}, being the group of rotations of a square. We must therefore decide if the right table is also isomorphic to C_{4}. First then, we note that *a* is not the identity element. But for this to be the Cayley table of a *group*, something must be! Looking at each row, we have that c is the identity. If this group is cyclic, we will have that this group is isomorphic to C_{4} (is this clear? Have a go. Cyclic will mean that every element is equal to *x ^{k}* for some special -though not necessarily unique- element

*x*). Just as with the group of rotations of an n-gon, there will be 2 possible generators. We note that, in the right table,

*a*does not generate the group. But both

*b*and

*d*will work. So let’s rearrange our table so that

*c*(the identity element) is the element of the first row and column and put

*b*next as it is a generator just like 90° in the left table, then

*a*as

*a*=

*b*. I’ll leave the left table how it was to be able to compare the tables. Before continuing: are the tables the ‘same’?

^{2}Although the change of symbolds is not as clear as the previous example (where we could just divide all of the entries of the left table by 90), these tables have the same structure. Before continuing: confirm this for yourself: how can you match the elements in the left table with those of the right table?

Ok, so pairing 0 with c, 90 with b, 180 with a, and 270 with d then means that the inside of the tables (the result of multiplying these elements) is the same. More formally we could express this as…

There are two subtle things here, that may not seem very clear until you’ve done more examples. First, we could have put *d* as the 2nd row and *b* as the 4th. This gives us an alternative pairing (writing out the tables and pairing explicitly will make this clearer). Secondly, if we decide the pair 90° and *b*, this restricts our other choices. Why? Well, if we need the inside of the tables to look the same (so that they have the same structure) then sending 90° to* b* means we’d need to send 180° to *a*. This would then define where 270° would need to go. Also, since the identity element is the only idempotent element (equal to its own square i.e. *e* is idempotent means *e*=*e ^{2}*) there is no choice on where the identity is sent: if it were sent anywhere else then the insides of the tables cannot match (I would recommend thinking about this).

We’ll do one more example. Consider these Cayley tables.

Before continuing: have a play. Are they the ‘same’? I think it’s always worth starting by working out what the identity element is and redrawing the table.

Spoilers! First of all, what’s the identity element? From the table, the element that ‘does nothing’ is clearly *z*. So we’ll redraw the table.

Study the table carefully. What is the inverse of each element?

The key thing to check is the orders of the elements given. For every element *g* in the right hand table, we have that *g ^{2}*=

*z*. This means that every (non-identity) element is of order 2. This means that these tables cannot be encoding the same group. For starters, the right hand group is not cyclic! (Why? Well, each element generates a group of size 2.) Sending 0° to

*z*is fine, but where can we send 90°? Trying the 3 possible choices

*w*,

*x*, and

*y*we can see that it won’t work! The right Cayley table is actually a direct product (seen in the last post). Have a go at working out what it actually is (answer in the comments).

**Summing up.** Hopefully these examples make you feel more confident with Cayley tables, but also show you that the Cayley table tells you everything about the structure of the group. We’ll come back to this next time, when I’ll finally get round to properly defining ‘same’!

I would recommend getting some more practice. There is a good page here detailing how to fill in an incomplete Cayley table. This can help you work towards the natural question: for small values of n, how many ‘different’ Cayley tables are there for a group of size n? Working on this challenge will give you lots of practice at working with Cayley tables. The answers can be found in the first table of this page. Best of luck!