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There are seven irreducible characters, one for each conjugacy class. Besides the trivial character and the four supplied rows, let the remaining degrees be $d,e$. The sum of squared degrees gives $d^2+e^2=360-(1+25+64+64+100)=106$, whose positive integer solutions are $\{d,e\}=\{5,9\}$. Complex conjugation permutes irreducible characters while preserving degree; the known rows are real and each unknown degree occurs just once among the remaining characters, so the two missing characters are real.

Call their entries $x_j,y_j$, with degrees five and nine. Orthogonality with the regular character gives $5x_j+9y_j=-{\sum }_{\mathrm{known}}\chi (1)\chi (C_j)$ for $j>1$. Column norms give $x_j^2+y_j^2=360/|C_j|-{\sum }_{\mathrm{known}}\chi (C_j{)}^2$. These equations yield two possible rational pairs in some columns. For this integral character reconstruction from column orthogonality, a further constraint is essential: character values are algebraic integers, so rational candidates must be integers. For example at $C_6$, $5x+9y=-9$ and $x^2+y^2=1$ allow $(0,-1)$ and $(-45/53,-28/53)$; the second pair is not integral. The integral choices complete the table:Every nontrivial row has $\chi (g)\ne \chi (1)$ for $g\ne 1$. In a unitary realization, $\chi (g)=\chi (1)$ exactly when all eigenvalues of the representing matrix are one, so each nontrivial irreducible representation is faithful. If a nontrivial proper normal subgroup existed, a nontrivial irreducible representation of the nontrivial quotient would pull back to a nontrivial representation with that subgroup in its kernel. This contradicts faithfulness. The group is simple.