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I am studying group theory and character table of $S_2$ is given in the book. But how to obtain this table is not given.

Can someone explain how exactly to construct this table?

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Once you have finished the book, you'll be able to fill in such tables yourself. And for S2, it will take like 30 seconds. –  Sjoerd Nov 20 '11 at 16:45
    
@Pratik: Your accounts will be merged soon; unfortunately there seems to be an issue that the SE developers have to deal with first though. –  Zev Chonoles Nov 27 '11 at 22:38
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1 Answer 1

S2 is a group with two elements, call them 1 and t. A representation of S2 takes 1 to the identity matrix and t to some matrix, so we can think of representations X of S2 as simply some matrix X(t). Of course not just any matrix will do, since X is a homomorphism and so $(X(t))^2 = X(t^2) = X(1)$ has to be the identity matrix. In other words, X(t) has to be a matrix whose square is the identity. All of its eigenvalues also have to square to the identity, so they are all ±1. By considering the Jordan canonical form of X(t), you can see that X(t) must be diagonalizable (since a large Jordan block does not square to the identity).

Of course if we diagonalize X(t), then each of the standard subspaces is an invariant subspace, so X is only irreducible if it is one dimensional.

However, there are not very many 1×1 matrices whose eigenvalues are just ±1. They are X1(t) = [1] and X2(t) = [-1]. These are both irreducible representations of S2, so we have found them all. Writing down their traces we get:

$$\begin{array}{r|rr} S_2 & 1 & t \\ \hline X_1 & 1 & 1 \\ X_2 & 1 & -1 \end{array}$$

Similar ideas work for any finite abelian group. One takes a minimal generating set and arbitrarily assigns appropriate roots of unity to the generators, giving one one-dimensional irreducible representation per element of the group.

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