Find character table of Klein four group
The representation of Klein four group is, $V=\{a,b\mid a^2=b^2=(ab)^2=e\}$
Now, to determine the character table, first I need to find out all the conjugacy class. There are four elements in $V$. But I don't see how to get the conjugacy classes from the representation. Hence, use permutation group elements, $V=\{e,(12),(34),(12)(34)\}$, to explicitly get some idea. Now, isn't there should be three conjugacy classes? (as per identity, two cycle and 4 cycle)
\begin{align*} \begin{array}{c | c| c | c } & h_1=1 & h_2=2 & h_3=1\\ \hline \chi^1 & 1 & 1 & 1\\ \hline \chi^2 & a & c & e\\ \hline \chi^3 & b & d & f\\ \end{array} \end{align*}
$$1^2+a^2+b^2=4\implies a^2+b^2=3$$
Then, either, $a=2,b=1$ or $a=1,b=2$. Picking the first choice, and then apply first and second orthogonality formula I can fill the rest. but when I googled the solution, I get the table is incorrect. It is $4\times 4$. It seems like the only issue I was facing to determine how to get conjugacy classes, if I have only given the representation. Is there any way to get that for smaller group?
As @testaccount pointed out, I messed up with the conjugacy classes with $S_4$. Fixing them I get,
\begin{align*} \begin{array}{c| c | c| c | c } & h_1=1 & h_2=1 & h_3=1 &h_4=1\\ \hline \chi^1 & 1 & 1 & 1 & 1\\ \hline \chi^2 & 1 & a & b & c\\ \hline \chi^3 & 1 & d & e & f\\ \hline \chi^4 & 1 & g & h & i\\ \end{array} \end{align*}
\begin{align} 1+a+d+g&=0,\\ 1+b+e+h&=0,\\ 1+c+f+i&=0 \end{align}, \begin{align} 1+a+b+c&=0,\\ 1+d+e+f&=0,\\ 1+g+h+i&=0 \end{align}
but now the system of equation seems very complicated to solve.
A conjugacy class of $x$ in $G$ is a set $\{gxg^{-1} : g \in G\}$. In the case of an abelian group $G$, we have $gxg^{-1} = x$ for all $g \in G$, so every conjugacy class is a singleton $\{x\}$. Therefore in your example, there are $4$ conjugacy classes in $V$.
You confusion might arise from the description of the conjugacy classes in the symmetric group, where conjugacy is determined by cycle structure.
Namely you write as a permutation group $V = \{(1), (12), (34), (12)(34) \}$. So $V < S_4$, where $S_4$ is the symmetric group on four letters.
While it is true that $(12)$ and $(34)$ are conjugate in $S_4$, it is not true that they are conjugate in $V$. (Again, because $V$ is abelian.) For example $\sigma (12) \sigma^{-1} = (34)$ for $\sigma = (13)(24) \in S_4$, but $\sigma \notin V$.