Given $G$ is a group of order 24 with the following table of characters
Find the value for element A.
These are all the instructions that are given. I guess I am not really getting this question. First I am assuming (without loss of generality) the G is an additive group , where the identity element is $0$ and the inverse of any $B \in G$ is $-B$. And I proceed as follows.
$$X_5 + C_3 = A \implies -C_3 - X_5 = -A$$
Since $X_5 + C_5 = 1$, and $X_4+C_5=1 \implies -C_5-X_4 = -1$
, then I can then write $$-C_3 - X_5 + X_5 + C_5 -C_5-X_4= -A + 1 - 1$$ $$\implies -C_3 - X_4 = -A \implies X_4 + C_3 = A = -1$$
However, to check the calculation I try using different elements, and I get a different conclusion. For example
$$X_5 + C_2 = -1 \implies -C_2 -X_5 = 1,\ \text{and}\ X_2 + C_2 = -1$$ and so $$X_2 + C_2 - C_2 - X_5 + X_5 + C_3 = -1 + 1 + A$$ $$\implies X_2 + C_3 = A = 1$$
So like I said it seems I am missing something very important here.
Recall that an Abelian group must have all of its irreducible characters of degree 1 and therefore you may not assume $G$ is additive since $\chi_3(C_1) =2,$ $ \chi_4(C_1) = 3,$ and $ \chi_5(C_1) = 3$ are respectively the degrees (dimensions of the vector spaces) corresponding to the $3^{rd},$ $4^{th},$ and $5^{th}$ irreducible representations respectively. This is because $C_1$ is the conjugacy class of the identity element and therefore $\rho(e) = \text{trace}(I_{\text{deg}(\rho) \times \text{deg}(\rho)}) = \text{deg}(\rho) = \text{dim}(V) $, where $\rho: G \longrightarrow V$ is a representation and $I_{k \times k}$ is the ${k \times k}$ identity matrix, see chapter 2 of Fulton & Harris or chapter 2 of Serre or even the wikipedia article on character tables. However using the Shur orthoganility relations we know that \begin{equation} \sum_{i}\chi_{i}(C_k)\chi_{i}(C_3) = 0 \end{equation} if $k \neq 3$ and therefore choosing say $k=1$ we get that \begin{equation} \sum_{i}\chi_{i}(C_1)\chi_{i}(C_3) = 1\cdot 1 + 1\cdot 1 + 2\cdot 2 + 3\cdot (-1)+3\cdot A = 1 + 1 + 4 - 3+ 3 A =0 \end{equation} and solving for $A$ we get that \begin{equation} A = \frac{-1}{3}(1+1+4-3) = -1. \end{equation}