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I recently saw a question in a Text book, which asks to prove that "The group of symmetries of the polynomial $x_1x_2 + x_3x_4 + x_5x_6$ is a subgroup of $S_6$ of order $48$".

(By the group of Symmetries of this polynomial, we mean the stabilizer of the polynomial $x_1x_2+x_3x_4+x_5x_6$ in the action of the group $S_6$ on $\mathbb{Z}[x_1,x_2,\cdots,x_6]$ given by $\sigma.f(x_1,x_2,\cdots,x_6)=f(x_{\sigma(1)},x_{\sigma(1)},\cdots,x_{\sigma(6)})$.)

In view of this exercise, I would like to ask the following question:

Is every finite group realizable as the full stabilizer of a polynomial over $\mathbb{Z}$ in a certain number of indeterminates? If yes, then how can we construct that polynomial?

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It seems to me this is related to the "inverse Galois problem", q.v., which asks whether every finite group is the Galois group of some extension of the rationals. –  Gerry Myerson Aug 9 '12 at 13:00
    
@Gerry And this question, which asks the same for Ideal class groups of Dedekind Domains. [1]: math.stackexchange.com/questions/101859/… –  Ravi Donepudi Aug 10 '12 at 18:00

1 Answer 1

up vote 7 down vote accepted

If $X$ is a graph, define the polynomial $p_X$ to be the sum of the monomials $x_ix_j$, where $ij$ runs over the edges of $X$. (The polynomial in the question is the polynomial of a perfect matching.) Then the automorphism group of this polynomial (in the sense defined above) is the automorphism group of the graph $X$. It has long been known that every finite group is the full automorphism group of a finite graph. For background, see the wikipedia article on Frucht's theorem.

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This method has the disadvantage of only working for abstract groups rather than permutation groups. For instance the alternating group on 4 points requires only 4 variables using general polynomials, but requires 10 variables using directed graphs (and somewhere between 12 and 24 for undirected graphs, I believe). –  Jack Schmidt Aug 9 '12 at 18:07

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