= Solution
Let $\phi=(1+\sqrt5)/2$ and $\bar\phi=(1-\sqrt5)/2$. In class order $(1,2A,3A,5A,5B)$, <character orthogonality> gives
$$
\begin{array}{c|rrrrr}
&1&2A&3A&5A&5B\\
1&1&1&1&1&1\\
3&3&-1&0&\phi&\bar\phi\\
3'&3&-1&0&\bar\phi&\phi\\
4&4&0&1&-1&-1\\
5&5&1&-1&0&0
\end{array}
$$
with class sizes $1,15,20,12,12$. The rows are orthonormal and their squared degrees sum to $60$, so the table is complete.
A three-dimensional representation of $S_5$ would decompose into irreducibles. Since $S_5$ has irreducible degrees $1,1,4,4,5,5,6$, only one-dimensional constituents could occur, whose restrictions to $A_5$ are trivial. It cannot restrict to either irreducible $3$.
Back to article page
