Am I able to calculate the character table of A4 without using the character table of S4?

Question

I've been taught how to determine the character of table of A4 using the character table of S4, but I was wondering if there is a way I could calculate the character table of A4 without using S4's character table?

Answer 1

Sure.

First of all, the derived subgroup $[A_4,A_4]$ of $A_4$ is $V_4=\{Id, (12)(34), (13)(24), (14)(23)\}$, and we have an group isomorphism $A_4/[A_4,A_4]\simeq \mathbb{Z}/3\mathbb{Z}$ (since the quotient has order $3$), which sends (123) to $1+3\mathbb{Z}$.

The general theory says that you have $3$ irreducible representations of degree $1$. Any other irreducible representation has at least degree $2$. But the only decomposition of $12$ as a sum of squares containing three $1$'s is $12=1^2+1^2+1^2+3^2$, so there is only one other irreducible representation, which has to have degree $3$.

You can also prove this fact by saying that there is exactly four conjugacy classes of $A_4$, represented by $Id, (12)(34), (123), (132)$.

The characters of a cyclic group of order $3$ are well known, and by inflation you get your three characters of $A_4$ corresponding to representations of degree $1$. This is not too difficult to compute the values of the inflated characters using the explicit isomorphism.

Now, you have the last row to fill in in the characters table (the one corresponding to the representation of degree $3$), but to do so, just use the orthogonality relations.