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: 
You can embed the Klein 4-group (which is not cyclic, since it contains no element of order $4$!) into $S\_4$ in four ways: 

$$\langle (12),(34)\rangle=\{1,(12),(34),(12)(34)\} $$

$$\langle (13),(24)\rangle=\{1,(13),(24),(13)(24)\} $$
$$\langle (14),(23)\rangle=\{1,(14),(23),(14)(23)\} $$
$$\langle(12)(34),(14)(23)\rangle=\{1,(12)(34),(14)(23),(13)(24)\}$$

but only the last subgroup happens to be normal.
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.