# Working with character tables

I am currently a bit stumped by an old exam question, which gives a character table and wants me to deduce properties of the group:

1. What is the order of $$g$$?
2. Show that $$g \notin C_G(G)$$
3. Show there is a $$h \in G$$ of equal order which is not conjugate of g
4. Show there are two irreducible representations of $$G$$ with the same dimensions
5. Show G has an irreducible representation of degree $$\geq 6$$

So far I have the following:

1. By the second orthogonality relation I get that $$|C_G(g)|=11$$ and since $$\langle g \rangle$$ is a subgroup of it, this only leaves 11 as the possible order of $$g$$.
2. $$g \in C_G(G)$$ would mean that every $$h \in G$$ is conjugate of $$g$$, but the character table has $$0$$ entries and characters are constant on conjugacy classes. Hence, if $$g \in C_G(G)$$ that would mean the corresponding character $$\chi$$ is $$0$$, a contradiction to the definition of the character table requiring $$\chi$$ to be irreducible.

Is this correct? How would I go about showing the other questions? I know the general process of creating a character table but this is my first time working with it this way.

Thanks!

Your solution to part 1 is correct. Part 2... not so much. Here is how I would approach these. Let $$\chi_i$$ denote the character corresponding to the $$i$$-th row of the table.

1. If $$g$$ is central, then $$|C_G(G)| \ge |g| = 11$$ implies that $$G$$ has at least 11 conjugacy classes, hence at least 11 irreps. But the table has just 10 rows.

2. One such element is $$h = g^{-1}$$. This is because $$\chi_6(g^{-1}) = \overline{\chi_6(g)} \neq \chi_6(g)$$.

3. Necessarily $$\chi_7 = \overline{\chi_6}$$. In particular, $$\chi_6(1) = \chi_7(1)$$.

4. The only possible characters of degree 1 are $$\chi_1$$ and $$\chi_9$$ since those are the only ones whose value at $$g$$ is an $$11$$-th root of unity. We know that $$\chi_6(1) = \chi_7(1)$$ from above, so $$\sum_i \chi_i(1)\chi_i(g) = \chi_1(1) - \chi_2(1) - \chi_4(1) - \chi_5(1) - \chi_6(1) + \chi_9(1) \le \chi_1(1) + \chi_9(1) - 8.$$ By orthogonality, the sum on the left-hand side is $$0$$, so $$\chi_1(1) + \chi_9(1) \ge 8$$. Either $$\chi_1$$ or $$\chi_9$$ is the trivial character, so one of these characters has degree $$\ge 7$$.