Assume we're given a group of order 10 with 4 conjugacy classes with representatives $g_i$ for $1\leq i\leq 4$ so that $(|C_G(g_i)|:1\leq \leq 4) = (10,5,5,2)$ where $C_G(g_i) := \{h \in G : g_i h = h g_i\}$. Two rows of the character table are given in the following table:
Find, with proof, the other rows of the character table.
I know I should use some properties of character tables such as the fact that the columns of the character table are orthogonal and hence the character table is invertible. This follows from the theorem that if $C,C'$ are conjugacy classes of G and $g\in C, h\in C',$ then $\sum_{i=1}^s \chi_i (g)\overline{\chi_i}(h) = |G|/|C|$ if $C=C'$ and 0 otherwise. But I'm not sure if this will help deduce the other rows of the character table.
Note that $C_G(g_2)$ has index $2$ in $G$ and is hence normal. Hence, $G$ admits the sign character $\chi_3$, i.e., the unique non-trivial character of the form $G \to G/C_G(g_2) \to \{\pm 1\}$. Since $\lvert C_G(g_3)\rvert = 5$, the only character $C(g_3) \to \{\pm 1\}$ is the trivial one. Thus, $\chi_3(g_3) = 1$. Clearly, $\chi_3(g_1) = \chi_3(g_2) = 1$. Since $\chi_3\neq \chi_1$, we must have $\chi_3(g_4) = -1$.
From the relation $10 = \lvert G\rvert = \sum_{i=1}^4 \chi_i(1)^2 = 1^2 + 2^2 + 1^2 + \chi_4(1)^2$, we deduce $\chi_4(1) = 2$. Thus, the character table has the form $$ \begin{array}{r|cccc} & g_1 & g_2 & g_3 & g_4 \\ \hline \chi_1 & 1 & 1 & 1 & 1 \\ \chi_2 & 2 & \frac{-1+\sqrt{5}}{2} & \frac{-1-\sqrt{5}}{2} & 0 \\ \chi_3 & 1 & 1 & 1 & -1 \\ \chi_4 & 2 & a & b & c \end{array} $$ Since the columns need to be orthogonal, we deduce $$ \begin{align} a &= \frac{-1-\sqrt{5}}{2} \\ b &= \frac{-1+\sqrt{5}}{2} \\ c &= 0. \end{align} $$