I’m not a great fan of bank holidays, because I generally don’t like the disruption of my normal routines, but one thing bank holidays are great for is getting ahead with my study schedule! This week I’ve had loads of time to wrap up TMA01, and to work through Unit GTA3, which is about permutations.

I’d hoped that there would be some crossover between the combinatorics stuff that we covered in MS221 Unit B1, and the group theory permutations material in GTA3. It turned out that although there weren’t any explicit connections made between the two topics, an understanding of the basics of permutations (in the sense of questions like “how many different ways can I arrange the numbers 1, 2, 3 and 4?”) was definitely useful to have.

I think the two best things about this unit, though, are the introduction of cycle form notation and Cayley’s Theorem. I really love cycle form, it’s just so much more compact and efficient than two-line symbols. For example, here’s a symmetry in two-line notation:

$\left( {\begin{array}{*{20}{c}}1 & 2 & 3 & 4 \\1 & 2 & 4 & 3 \\\end{array}} \right)$

And here’s the same symmetry in cycle form:

$\left( {\begin{array}{*{20}{c}}3 & 4 \\\end{array}} \right)$

I love the fact that you don’t have to write down any symbols that are fixed, such as the 1 and 2 in the symmetry given above.

I have to say, Cayley’s Theorem really blew my mind. Every finite group is isomorphic to a permutation group, and it’s staggeringly easy to write down the related permutation group, as long as you’ve got the Cayley table for the finite group. For example, here’s the Cayley table for an arbitrary group $\left( {\left\{ {e,a,b,c} \right\}, \circ } \right)$:

$\circ$ e a b c
e e a b c
a a c e b
b b e c a
c c b a e

If we want to find a permutation group which is isomorphic to this group, we can just couple the top row of the Cayley table with each of the rows below in turn, to give us a set of four permutations in two-line notation, like this:

$\left\{ {\left( {\begin{array}{*{20}{c}} e & a & b & c \\ e & a & b & c \\\end{array}} \right),\left( {\begin{array}{*{20}{c}} e & a & b & c \\ a & c & e & b \\\end{array}} \right),\left( {\begin{array}{*{20}{c}} e & a & b & c \\ b & e & c & a \\\end{array}} \right),\left( {\begin{array}{*{20}{c}} e & a & b & c \\ c & b & a & e \\\end{array}} \right)} \right\}$

Then once the permutations are in two-line form, we can convert them in cycle form:

$\left\{ {e,\left( {e,a,c,b} \right),\left( {e,b,c,a} \right),\left( {e,c} \right)\left( {a,b} \right)} \right\}$

How awesome is that? Every finite group can be given the same treatment, and will end up as a nice, (relatively) compact permutation group in cycle form. Hooray for Cayley!