1. ## Character/ Representation Theory

Can anybody help me with these questions as I am completely lost and haven't done any character theory or rep theory for a long time

(1) Let X be the matrix comprising the character table of a group.
Show that X is non-singular & hence prove that the number of real-valued characters is equal to the number
of self-inverse conjugacy classes.

(2) Show that in a group G of odd order, no element other than the identity
is conjugate to its inverse. Deduce that in a group of odd order, the trivial character is the only
irreducible character that is real-valued.

2. Originally Posted by Turloughmack
Can anybody help me with these questions as I am completely lost and haven't done any character theory or rep theory for a long time

(1) Let X be the matrix comprising the character table of a group.
Show that X is non-singular & hence prove that the number of real-valued characters is equal to the number
of self-inverse conjugacy classes.
This is a sketch of the proof borrowed from the book "Representations and characters of groups" by Liebeck (p264).

The irreducible character of G are linearly indepdent over $\mathbb{C}$ as functions from G to $\mathbb{C}$, which follows that X is non-singular.

Let $\bar{X}$ be the complex conjugate of matrix X. For each irreducible character $\chi^i$, it is known that $\bar{\chi^i}$ is also an irreducible character. Thus, we can construct a permutation matrix P induced from $\chi^i \rightarrow \bar{\chi^i}$ such that $PX=\bar{X}$.

Similarly, for each conjugacy class $C_i(g_i)$, the entries in the column of X corresponding to $C_i(g_i)$ are the complex conjugates of the entries in the column of X corresponding to $\bar{C_i}(g_i^{-1})$, where $C_i$ denotes the conjugacy class of its representative $g_i$, and $\bar{C_i}$ the conjugacy class of $g_i^{-1}$. Similarly, we can construct a permutation matrix Q induced from $C_i \rightarrow \bar{C_i}$ such that $XQ=\bar{X}$. It follows that $P=XQX^{-1}$. Thus, we see that P and Q have the same trace. Recall that trace of a permutation matrix corresponds to the number of fixed points by the permutation. Therefore, the number of irreducible real-valued characters (corresponding to the fixed points of the first permutation) is the trace of P, and the number of self-inverse conjugacy classes (corresponding the the fixed points of the second permutation) is the trace of Q, which are equal.

(2) Show that in a group G of odd order, no element other than the identity
is conjugate to its inverse. Deduce that in a group of odd order, the trivial character is the only
irreducible character that is real-valued.
Note that $C_i=\bar{C_i}$ iff each $g_i$ is conjugate to $g_i^{-1}$, i.e., $hg_ih^{-1}=g_i^{-1}$ for some $h \in G$. Then, $h^2g_ih^{-2}=g_i$, $h^3g_ih^{-3}=g_i^{-1}$, etc. Since G is of odd order, we see that $g_i=g_i^{-1}$(verify this). Thus, only the identity is conjugate to its inverse. Hence, the self-inverse conjugacy class is {1}. It follows that the number of irreducible real-valued characters is only one by (1).

3. Alternatively, one can appeal to the amazing theorem that for a finite group $G$ if $\sqrt{g}=\#\left\{h\in G:h^2=g\right\}$ then the number of self-conjugate representations of $G$ is $\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}$. Although, the proof of this comes startling close to the ideas that the previous paster mentioned.