Consider the polynomial $f(x_1,x_2,x_3,x_4) = x_1 x_2 + x_3 x_4.$ Let $G$ denote the symmetry group of $f$ consisting of those permutations of $x_1,x_2,x_3$ and $x_4$ which leave $f$ fixed. Show that $G \cong D_4$ where $D_4$ is the symmetry group of the square.
I have found the symmetry group of $f$ to be $H=\{(1),(1\ 2),(3\ 4), (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3), (1\ 3\ 2\ 4), (1\ 4\ 2\ 3) \}.$ I know that $D_4 \cong G$ where $G$ is a group generated by two of its elements $a$ and $b$ such that $o(a)=4$ and $o(b)=2$ along with $ba=a^3 b.$ From this stage how should I proceed? Do I take $a=(1\ 3\ 2\ 4)$ and $b=(1\ 2)$ or $(1\ 3)(2\ 4)$ here or should I go for other options?
As I am new to this subject I dont know how to conclude.Please help me.
Thank you in advance.
Since they specifically meantion squares, lets' get geometric. Draw a square and put your variables in the corners in the following pattern: $$ \begin{matrix} x_1&x_3\\x_4&x_2 \end{matrix} $$ Then $f$ is the sum of the product of the two diagonals. Surely, applying a symmetry of that square preserves which variables are diagonally opposite one another. Thus $D_4$ is a subgroup of $G$.
Conversely, any permutation that preserves which elements are diametrically opposite one another is a symmetry of that square: If I make a different square using the same numbers with $x_1$ and $x_2$ diametrically opposite one another, then there is an element of $D_4$ that translates between my two squares. Thus $G\subseteq D_4$.