Finding square of an element from a character table I'm trying to compute the alternating and symmetric squares of a $\chi_2$, from this incomplete character table:
$$
\begin{array}{c|ccccccc}
g_i & g_1&g_2&g_3&g_4&g_5&g_6&g_7 \\
|C_G(g_i)|&360&8&9&9&4&5&5\\
\hline
\chi_1 &1&1&1&1&1&1&1\\
\chi_2&5&1&2&-1&-1&0&0\\
\chi_3&5&1&-1&2&-1&0&0
\end{array}
$$
I am using the formulas 
$$\chi_S(g_i)=\frac{1}{2}(\chi^2(g_i)+\chi(g_i^2))$$
and
$$\chi_A(g_i)=\frac{1}{2}(\chi^2(g_i)-\chi(g_i^2))$$
and have been given the following squares:
$$g_5^2\in g_2^G,g_6^2\in g_7^G,g_4^2\in g_4^G,\text{ and }g_3^2\in g_3^G.$$
Clearly $g_1^2=g_1$, and I can see that $g_7^2$ is either in the conjugacy class $g_6$ or $g_7$ (since the order of $g_7$ must divide $|C_G(g_7)|=5$, and cannot be $1$). It doesn't matter for my purposes which it is in, since $\chi_2(g_6)=\chi_2(g_7)$.
However, I was only able to obtain that $g_2^2 = g_1$ by means of trial-and-error; calculating the inner products $\langle \chi_S,\chi_S\rangle$ and $\langle \chi_A,\chi_A\rangle$ and ensuring that they are both integers. It seems this might not always give a unique answer for some tables. Is there a better way of seeing this from the information given? (I know that the group is $A_6$, but this is not given in the question).

Since $g_5 \in C_G(g_5)$, the order of $g_5$ is either $2$ or $4$. Consequently, $g_5^2$ has order $1$ or $2$. Since $g_5^2 \in g_2^G$, the order of $g_2$ is either $1$ or $2$. Since $g_2 \neq g_1$, the order of $g_2$ is $2$ and so $g_2^2 = g_1$. 
