= Character table of the alternating group A5
{c}
{title2=$A_5$}
In the <conjugacy class> order $1,(123),(12)(34),(12345),(13452)$, the five complex <irreducible characters> of the <alternating group> $A_5$ are
$$
\begin{array}{c|rrrrr}
&1&(123)&(12)(34)&(12345)&(13452)\\\hline
1&1&1&1&1&1\\
3&3&0&-1&(1+\sqrt5)/2&(1-\sqrt5)/2\\
3'&3&0&-1&(1-\sqrt5)/2&(1+\sqrt5)/2\\
4&4&1&0&-1&-1\\
5&5&-1&1&0&0
\end{array}.
$$
The class sizes are $1,20,15,12,12$. Restricting the degree-four and degree-five <irreducible characters> of $S_5$ gives rows of squared <character inner product> one. The degree-six restriction has squared norm two and is orthogonal to those rows and the trivial row, so it splits into the two remaining irreducibles. The sum of squared degrees forces their degrees to be three. Orthogonality forces their last two values to have sum one and product minus one, producing the displayed roots. Since a <normal subgroup> is a union of classes and its order divides sixty, these class sizes also prove the <Simplicity of the alternating group A5>.
Back to article page
