# Permutation problem from Alexandroff group theory text book

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.

• Just a comment : "addition table" is not really the appropriate term. Usually we use "addition" for abelian groups, and "multiplication" for groups. – user171326 Aug 31 '16 at 22:24

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

$$|H|=\frac{|S_4|}{6}=\frac{24}{6}=4.$$

You're looking for permutations which leaves invariants the polynomial $x_1x_2 + x_3 + x_4$.

An example of such permutations is $\sigma = (1,2)(3,4)$.

In fact, $\sigma (x_1x_2 + x_3 + x_4) = (x_2x_1 + x_4 + x_3) = x_1x_2 + x_3 + x_4$.

Can you find all such $\sigma$ ?

Let's start with an example of a permutation that fails to preserve the polynomial: the transposition $(2 \; 3)$. It maps the polynomial to $$x_{1} x_{3} + x_{2} + x_{4},$$ a different one.

This (and perhaps experimentation with a few more such examples) will convince you that the only permutations that preserve the polynomial are the ones that, for each variable, preserve its degree in every monomial that contains this variable. For example, the monomial $x_{1}^3 x_{2}$ would be preserved only by the trivial group.

Namely, for your polynomial, the only such permutations are the transpositions $(1 \; 2)$ and $(3 \; 4)$, which together generate the Klein four-group https://en.wikipedia.org/wiki/Klein_four-group, which is your $H$.

• Typo : the transpositions should be $(1 2)$ and $(3 4)$. – user171326 Aug 31 '16 at 22:30
• Why? If we interchange $x_{1}$ and $x_{2}$, that changes the polynomial to $x_2 x_3 + x_1 + x_4$. – avs Aug 31 '16 at 22:36
• The OP's polynomial is $x_1x_2 + x_3 + x_4$, isn't ? – user171326 Aug 31 '16 at 22:39
• Also, the transposition $\tau = ( 3 4)$ is non-trivial but $x_1^3x_2$ is invariant under $\tau$. – user171326 Aug 31 '16 at 22:45
• My bad: looked at the wrong polynomial. I will correct the answer. With my monomial example, I just meant to have the two variables $x_1, x_2$. Sorry I didn't make that clear. – avs Aug 31 '16 at 22:55