 Solution

Put $w_j=|C_j|/|G|$. The character orthogonality for the five supplied rows say ${\sum }_j{w}_j{\chi }_r(C_j){\chi }_s(C_j)={\delta }_{rs}$. Solving these linear equations givesFor example orthogonality of the two degree-four rows gives $16w_1+w_3+w_4-4w_5-w_6=0$, and their norms give $16w_1+w_3+w_4+4w_5+w_6=1$; the remaining row products determine the other weights. Since $C_1=\{e\}$, $w_1=1/|G|$, so $|G|=120$ and the class sizes are the displayed numerators.

Tensoring the fifth irreducible representation with the second, one-dimensional one gives another irreducible character, with row $(5,1,-1,0,-1,-1,1)$. It differs from all five known rows. There are seven conjugacy classes, hence seven irreducible characters. The sum of squared degrees is $|G|$, so the final degree is $\sqrt{120-(1+1+16+16+25+25)}=6$.

The regular representation has character zero off the identity and is the sum of irreducible characters weighted by their degrees. Applying this column by column determines the last row. The completed character table, in the supplied class order, isNo identification of the group with a familiar group is needed.