character table of group I have some problem with writing character table of a group. For instance, a group $S_4$. When we write character table, we write irreducible representations of group. So, how can I quickly find them? Then how to Fill the table? Can someone explain me upon this example?
 A: Let's start with $S_3$.
Step 1: Find the conjugacy classes when this is not too difficult. $1, (12), (123)$ generate the full group.
Step 2: There are two easy representations: the trivial one, and the "alternating" representation, which is just $\mathrm{sgn}$ which assigns to a cycle its parity in the decompmosition into transpositions. Thus, we get values $(1,-1,1)$ on the three conjugacy classes respectively.
Step 3: Finally, we have the permutation representation $S^3 \to GL(\mathbb C^3)$ which basically acts by permutation of indeces on a $3$-tuple. However, this representation decomposes into the trivial representation on the diagonal $\mathrm{Span}[1,1,1]:=U$ and its orthogonal complement. Hence, we gat that $\mathbb C^3:=U \oplus V$. But the values of the permutation representation should be $(3,1,0)$ on each conjugacy class (check this by looking at matrices.) Hence, the two dimensional irreducible $V$ representation has character $\chi_{\mathbb C^3}-\chi_{U}=(3,1,0)-(1,1,1)=(2,0,-1)$ which is the last character, of dimension $2$.
We know we are done since the sum of the squares of dimension is the cardinality of $S_3$, which is $6$.
Hint for $S_4$: Use the "same" $3$ representations and use that $\sum \mathrm{dim}\,\chi_i^2=24$, while there will be at most $5$ irreducible characters. Try tensoring for an algebraic way to get another one, and the last one can be deduced just for orthogonality reasons (use the inner product.)
If you get stuck, I suggest reading section $2.3$ of Fulton & Harris.
A: You want to find the character table of a small finite group. The quickest way is to look them up in a book or website, or use a program like GAP. Failing that, you can find it for yourself with a general method. There are a lot of shortcuts depending on the particular group. We could use orthogonality properties and other properties of characters. I suggest first finding the conjugacy classes and then number them. For example  $C_1$ is the conjugacy class of the identity. For $S_4$ there are four others.
We now identify a representative element of each conjugacy class. $C_1$ has the identity permutation $[1,2,3,4]$ as its only element. Let $C_2$ be the class of
permutation $d:=abab=[2,1,4,3]$ with $3$ elments of order $2$. Let $C_3$ be the class of permutation $a:=[1,3,4,2]$ with $8$ elements of order $3$. Let $C_4$ be the class of permutation $b:=[4,2,3,1]$ with $6$ elements of order $2$. Let $C_5$ be the class of permutation $c:=ab=[4,3,1,2]$ with $6$ elements of order $4$. A total of $24$ group elements.
Now the fun part. We define a multiplication of conjugacy classes as multisets of group elements. That is, given two multisets of group elments, the product is the multiset of all pairwise products of elments of the two multisets. Thus, $C_1$ is also the identity for conjugacy class multiplication. Now $C_2C_2=3C_1+2C_2$ because the $3$ elements of order $2$ gives the $3C_1$ where we multiply an element by itself, and for the product of two distinct elements of $C_2$ we get an element of $C_2$ in two ways.
Now we get a system of ten equations in the conjugacy classes and also require $C_1=1$. We got $C_2C_2=3C_1+2C_2$. Next, $C_3C_2=3C_3, C_3C_3=8C_1+8C_2+4C_3$. There are eight more equations. The last is $C_5C_5=6C_1+2C_2+3C_3$. Solving the system of equations gives five solutions, namely, $[1,3,8,6,6], [1,3,8,-6,-6], [1,3,-4,0,0], [1,-1,0,2,-2], [1,-1,0,-2,2].$ These solutions are almost the irreducible characters of our group, but we make two adjustments.
First, divide each solution value of $C_n$ by $c_n$, the number of elements of class $C_n$. This gives us $[1,1,1,1,1], [1,1,1,-1,-1], [1,1,-1/2,0,0], [1,-1/3,0,1/3,-1/3], [1,-1/3,0,-1/3,1/3].$ Now, we scale each vector so that the sum of $c_n$ multiplied by the squares of each value is $24$, the number of elements of the group. This scale factor turns out to be the dimension of each character representation. In our case they are $\{1,1,2,3,3\}$ respectively. Thus the characters are: $[1,1,1,1,1],[1,1,1,-1,-1],[2,2,-1,0,0],[3,-1,0,1,-1],[3,-1,0,-1,1].$
This method is completely computational and can be quickly and easily done (for small groups) by computer algorithms. You don't need to know any theorems in representation theory. The hardest part is solving the class multiplication equations, and fortunately, this can be easily done by using computer algebra systems in one command. There are other methods that require knowing special properties of group characters and special knowledge about the group in question.
