Monday 2 April 2007

Burnside's lemma (2)

Let's go on with Burnside's lemma. I want to show an example today. The question is how many different dichlorobenzenes there are. Of course for this question you don't need Burnside's lemma but I have to start somewhere.

If we don't take reflexions (because of chirality) then we have to consider the D6 point group. Let's call it G. X is the set of the 15 fixed dichlorobenzene structures seen in the last post.

What we are doing now is look at all the operations g in G and see how many structures in X stay the same when g is applied. Then you have to average that. The average happens to be the number of isomeric structures.

If you apply the identity E, everything stays the same and you have 15 structures fixed by the identity. There is no structure that stays the same when a 60° rotation (C2) or a 120° rotation (C3) are applied. With a 180° rotation around the main axes the 3 para-structures stay the same. For the two sets of 3 rotations perpendicular to the main axes (C'2 and C''2), 3 structures each stay the same.

This table summarises this.
n g|Xg| n |Xg|
1 E
15 15
2 C6 0 0
2 C3 0 0
1 C2 3 3
3 C'2 3 9
3 C''2 3 9


In the last line there is |G| to the left and the sum of the |Xg| to the right. Divide these two according to the formula I showed last time and you get 3 which is the true result: 0, m, p.

I hope I could kind of explain what's going on. It just seems to me that Burnside's lemma is some nice mathematics beyond the typical calculations.

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