Question: Let $\lambda$ and $\lambda\_L$ be the values of the function $f(x,y)$ at the optimum for problems 
\begin{align}
\lambda=\max\_{x}\min\_{y}f(x,y) 
\end{align}
\begin{align}
\lambda\_L=&\max\_{x}\min\_{y} f(x,y) \\& \mbox{s.t.}~~g(y)=c
\end{align}
respectively. We can assume that $x$ and $y$ comes from the same domains in both problems. Is $\lambda\geq \lambda\_L$?

Answer: 
One way to solve this problem is to use the left regular representation. $V$ acts on itself by left multiplication. First, we choose a numbering of the elements of $V$. Somewhat arbitrarily, I choose to label $1,a,b,ab$ by $1,2,3,4$, respectively. First note that $1$ must act as the identity permutation. We observe that $a$ acts by left multiplication by sending
\begin{align\*}
a: 1 &\mapsto a\\
a &\mapsto a^2 = 1\\
b &\mapsto ab\\
ab &\mapsto b \, .
\end{align\*}
Recalling our numbering we see that the permutation $\sigma\_a$ corresponding to $a$ sends
\begin{align\*}
\sigma\_a: 1 &\mapsto 2\\
2 &\mapsto 1\\
3 &\mapsto 4\\
4 &\mapsto 3
\end{align\*}
so $\sigma\_a = (1\ 2)(3\ 4)$. Proceeding similarly with $b$ and $ab$, we find $\sigma\_b = (1\ 3)(2\ 4)$ and $\sigma\_{ab} = (1\ 4)(2\ 3)$.

Note that choosing a different numbering for the elements of $V$ will yield different permutations. In fact changing the numbering conjugates all the permutations.
An easy way to do this is just to examine the permutations of order 2, since all elements of the Klein-4 group have order 2. These are just the set of all transpositions and disjoint transposition.

Note that the transpositions alone do not form a normal subgroup, so we have all of the pairs of disjoint transpositions: 
$$\{(),(12)(34),(13)(24),(14)(23)\}.$$