How to find an element fixed by a given group?

In Dummit's 1991 paper he presents an element $$\theta \in \mathbb{Q}(x_1, x_2, x_3, x_4, x_5)$$ that, when you ponder it, can be seen to be invariant to the 2 permutations $$\{(1 \, 2 \, 3 \, 4 \, 5), (2 \, 3 \, 5 \, 4)\}$$ that he says generate F20: $$\theta = x_1^2x_2x_5 + x_1^2x_3x_4 + x_2^2x_1x_3 + x_2^2x_4x_5 + \dots$$

He does not explain how he came upon that element and this is my question.

Is there a systematic way (perhaps using a comp algebra system to help with the calculations) to find some element in one of these sorts of extensions that is invariant to the permutations of a given group?

1 Dummit, D. S. “Solving Solvable Quintics.” Mathematics of Computation, vol. 57, no. 195, 1991, pp. 387–401.

• These permutations generate the Galois group of $X^5-2$, but whether this is useful or not ... with a CAS it should be easy to find. Commented Mar 17, 2023 at 16:12

Might as well promote my comments to an answer. This is somewhat speculative (I haven't looked at Dummit's paper), and also somewhat ad hoc. But it seems to fit.

I assume that this polynomial is well known to people studying Galois groups of quintics in general and $$F_{20}$$ in particular.

The 5-cycle $$\alpha=(12345)$$ and the 4-cycle $$\beta=(2354)$$ indeed generate a copy of $$F_{20}$$. You can quickly check that $$\beta\alpha\beta^{-1}=\alpha^2$$, so the group is a semidirect product $$C_5\rtimes C_4$$ with $$C_4$$ acting as the full group of automorphisms of $$C_5$$ (aka the holomorph of $$C_5$$).

Presumably Dummit needs a homogeneous polynomial in $$\Bbb{Q}[x_1,x_2,x_3,x_4,x_5]$$ that is invariant under $$F_{20}$$ but is NOT invariant under all of $$S_5$$. One tool for locating such polynomials is the Molien series, which is a generating function telling us the dimensions of the spaces of homogeneous invariant polynomials of a chosen degree.

To use the Molien series we need a tally of the cycle decompositions of the elements of the group we are interested in. The group $$F_{20}$$ has, in addition to the identity permutation, $$4$$ 5-cycles (powers of $$\alpha$$), $$10$$ 4-cycles (the conjugates of $$\beta$$ and its inverse), and $$5$$ products of two disjoint 2-cycles (the conjugates of $$\beta^2$$). Therefore the Molien series of $$F_{20}$$ is $$M(t)=\frac1{20}\left(\frac1{(1-t)^5}+\frac4{1-t^5}+\frac{10}{(1-t)(1-t^4)}+ \frac5{(1-t)(1-t^2)^2}\right)$$ Using Mathematica or any CAS of your choice it is easy to expand this as a Taylor series. Listing enough low degree terms: $$M(t)=1+t+2t^2+3t^3+6t^4+\cdots.$$ The coefficient of $$t^n$$ in $$M(t)$$ gives the dimension of the space of homogeneous degree $$n$$ invariants, so we see that:

• The space of linear homogeneous invariants has dimension $$1$$, obviously spanned by the first symmetric polynomial $$e_1=\sum_i x_i$$.
• The space of quadratic homogeneous invariants has dimension $$2$$, obviously spanned by $$e_1^2$$ and $$e_2=\sum_{i.
• The space of cubic homogeneous invariants has dimension $$3$$, obviously spanned by $$e_1^3$$, $$e_1e_2$$ and $$e_3=\sum_{i. So far no difference between $$F_{20}$$ and the fully symmetric polynomials.
• The space of quartic homogeneous invariants has dimension $$6$$. Here we get a difference because the space of $$S_5$$-invariants is spanned by $$e_1^4$$, $$e_2e_1^2$$, $$e_2^2$$, $$e_3e_1$$ and $$e_4$$ — only five generators.

So we know to look for a quartic. A way of finding candidates is to start with a monomial, and then let the desired group of permutations act on the subscripts of the variables, when the sum of the monomials in the orbit will automatically be an invariant. The group $$F_{20}$$ acts on the set of indices $$\{1,2,3,4,5\}$$ doubly transitively. Therefore it is of no use to us to consider a monomial involving only two variables — we automatically get all the variants of the said monomial under $$S_5$$. But a monomial of the type $$x_i^2x_jx_k$$, $$i,j,k$$ distinct, may fit our needs. There are $$\binom 51\binom 42=30$$ such monomials altogether. This is promising in the sense that they cannot be a single orbit of a group of order $$20$$!

As $$\beta^2=(25)(34)$$, it follows that the monomial $$P=x_1^2x_2x_5$$ is stabilized by $$\beta^2$$. So the sum $$S$$ of the monomials in the $$F_{20}$$-orbit of $$P$$ has only the following ten terms: $$x_1^2x_2x_5+x_1^2x_3x_4+x_2^2x_3x_1+x_2^2x_4x_5+x_3^2x_4x_2+x_3^2x_5x_1+x_4^2x_5x_3+x_4^2x_1x_2+x_5^2x_1x_4+x_5^2x_2x_3.$$ This seems to be the invariant Dummit is using.

The monomial $$x_1^2x_2x_3$$ has a trivial stabilizer in $$F_{20}$$, and hence its orbit accounts for the other $$20$$ monomials of the type $$x_i^2x_jx_k$$. If we denote the sum of the monomials in that bigger orbit by $$S'$$, we see that only the sum $$S+S'$$ is an invariant under all the permutations of the indices.

Acknowledgement: I am indebted to David E. Speyer for describing this use of Molien series to me in a related thread here in Math.Stackexchange. I am to be blamed for the quick and dirty ad hoc search method. A more knowledgeable user may be able to say something more definite.

• Thank you Jyrki, thanks for taking the time to write all that. Commented Mar 21, 2023 at 14:21