Exercise 3.1. Prove that if \(G\) is a perfect group and \(\chi\) is a complex character of \(G\) with \(\chi(1)=2\), then \(\chi(g)=2\) for all \(g\in G\), i.e., \(\chi\) is the sum of two trivial characters.

Exercise 3.2. Let \(p\) be a prime and let \(G\) be the group \(C_p\cdot C_{p-1}\), the normalizer of a Sylow \(p\)-subgroup of \(S_p\). Thus an element of order \(p-1\) in \(G\) acts on a normal subgroup of order \(p\) in \(G\) with trivial kernel.

1. (i) Prove that \(G\) has \(p\) conjugacy classes.
2. (ii) Write down the character table for \(G\).

Exercise 3.3. Let \(G\) be a finite group, let \(g\in G\) and let \(\rho\) be a representation of \(G\).

1. (i) Prove that the eigenvalues of \(g\rho\) are determined uniquely by the character values \(\chi(g^i)\) for all \(i\), where \(\chi\) is the character afforded by \(\rho\).
2. (ii) If \(\rho\) is the regular representation, prove that the eigenvalues of \(g\rho\) are all \(o(g)\)th roots of unity, each appearing with the same multiplicity.

Exercise 3.4. Here we give a completely different way to calculate the character table of \(A_5\), that relies on nothing much at all, just Exercise 3.1 and Exercise 3.3. Let \(G\) be a simple group of order \(60\).

1. (i) Prove that the irreducible character degrees of \(G\) are one of \(1,3,5,5\), \(1,3,3,4,5\) or \(1,3,3,3,4,4\).
2. (ii) Let \(x\in G\) has order \(5\). Using the \(1\)-eigenspace of \(x\) on the regular representation, prove that the first and third possible sets of character degrees are not possible. (Recall that if \(\chi(1)=n\) then there are \(n\) copies of \(\chi\) in the regular character.)
3. (iii) Determine how the eigenvalues of the regular representation on \(x\) are distributed among the irreducible characters, which have degrees \(1,3,3,4,5\).
4. (iv) Deduce that \(x\) is conjugate to \(x^{-1}\) but not to \(x^2\), and therefore conjugacy classes representatives are \(1,t,y,x,x^2\), where \(o(t)=2\) and \(o(y)=3\).
5. (v) Do the same for an element \(y\).
6. (vi) At this point you should have all columns except for that labelled by \(t\). Fill in the remaining column using orthogonality relations.

Exercise 3.5. Here is the character table of a finite group \(G\).

	\(x_1\)	\(x_2\)	\(x_3\)	\(x_4\)	\(x_5\)	\(x_6\)	\(x_7\)	\(x_8\)	\(x_9\)	\(x_{10}\)	\(x_{11}\)
\(\chi_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(1\)
\(\chi_2\)	\(14\)	\(-2\)	\(2\mathrm{i}\)	\(-2\mathrm{i}\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(1\)	\(1\)	\(1\)
\(\chi_3\)	\(14\)	\(-2\)	\(-2\mathrm{i}\)	\(2\mathrm{i}\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(1\)	\(1\)	\(1\)
\(\chi_4\)	\(35\)	\(3\)	\(-1\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(\eta_1\)	\(\eta_2\)	\(\eta_4\)
\(\chi_5\)	\(35\)	\(3\)	\(-1\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(\eta_2\)	\(\eta_4\)	\(\eta_1\)
\(\chi_6\)	\(35\)	\(3\)	\(-1\)	\(-1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(\eta_4\)	\(\eta_1\)	\(\eta_2\)
\(\chi_7\)	\(64\)	\(0\)	\(0\)	\(0\)	\(-1\)	\(1\)	\(1\)	\(1\)	\(-1\)	\(-1\)	\(-1\)
\(\chi_8\)	\(65\)	\(1\)	\(1\)	\(1\)	\(0\)	\(\theta_1\)	\(\theta_2\)	\(\theta_3\)	\(0\)	\(0\)	\(0\)
\(\chi_9\)	\(65\)	\(1\)	\(1\)	\(1\)	\(0\)	\(\theta_3\)	\(\theta_1\)	\(\theta_2\)	\(0\)	\(0\)	\(0\)
\(\chi_{10}\)	\(65\)	\(1\)	\(1\)	\(1\)	\(0\)	\(\theta_2\)	\(\theta_3\)	\(\theta_1\)	\(0\)	\(0\)	\(0\)
\(\chi_{11}\)	\(91\)	\(-5\)	\(-1\)	\(-1\)	\(1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(0\)	\(0\)


(Here, \(\zeta_i\) is a primitive \(i\)th root of unity, \(\eta_i=-\zeta_{13}^i-\zeta_{13}^{5i}-\zeta_{13}^{-5i}-\zeta_{13}^{-i}\), and \(\theta_i=\zeta_7^i+\zeta_7^{-i}\).)
1. (i) Why is \(G\) simple?
2. (ii) Prove that \(|G|=29120=2^6\cdot5\cdot7\cdot13\) and determine \(|C_G(x_i)|\) for \(1\leq i\leq 5\). Deduce that \(o(x_5)=5\), \(o(x_2)=2\) and \(o(x_3)=o(x_4)=4\).
3. (iii) Prove that, if \(P\in \mathrm{Syl}_2(G)\), that \(P\) is non-abelian, that \(Z(P)\) is elementary abelian, and that all involutions in \(Z(P)\) are conjugate in \(G\).
4. (iv) Prove that \(o(x_6)=o(x_7)=o(x_8)=7\) and \(o(x_9)=o(x_{10})=o(x_{11})=13\).
5. (v) It is known that the Sylow \(2\)-subgroups of \(G\) are trivial intersection, i.e., that \(P\cap P^x\) is either \(P\) or \(1\), for \(P\in\mathrm{Syl}_2(G)\) and \(x\in G\). Deduce that \(Z(P)\) has order \(8\), there are eight elements of order \(4\) squaring to a given involution in \(G\), and that \(P'=Z(P)\).

(\(G\) is the Suzuki group \(\mathrm{Sz}(8)\), one of an infinite series of simple groups discovered by Suzuki in 1960, special because they are the only simple groups for which \(3\) does not divide their order.)