I have the following question from Introduction to the Theory of Groups by Alexandroff. If somebody could please point the way to answering this question, rather than answering it directly that would be ace. I feel I could probably answer this if I understood more about what the question was saying - I'm a bit stumped and it seems quite a bit harder than earlier problems:
Prove that the permutations
\begin{pmatrix} 1 & 2 & 3 & 4\\ a_1 & a_2 & a_3 & a_4 \\ \end{pmatrix}
leaving invariant the polynomial $x_1x_2+x_3+x_4$, i.e. for which $x_{a_1}x_{a_2}+x_{a_3}+x_{a_4}$ is identical with $x_1x_2+x_3+x_4$, form a subgroup H of order 4 of the symmetric group $S_4$, and write down its addition table.
(H is called the group of the polynomial $x_1x_2+x_3+x_4$. A polynomial in $x_1$,$x_2$, $x_3$, $x_4$ whose group is $S_4$ is called symmetrical.
Let $S_4$ act on the set of polynomials in $4$ variables by permuting the variables (for example, $(2 3) \circ x_1 x_2+x_3+x_4=x_1x_3+x_2+x_4$). Then you are looking for the stabilizer of $p := x_1x_2+x_3+x_4$. It is a general fact of actions that the stabilizer of an element forms a subgroup. So the first claim follows immediately. Now we know an element in the orbit of this polynomial is completely determined by its $x_i x_j$ term. So the number of elements in the orbit of $p$ is $\binom{4}{2}=6$. By the orbit-stabilizer theorem we have
$$|H|=\frac{|S_4|}{6}=\frac{24}{6}=4.$$