A system of polynomials has Galois group G, a subgroup of $P_n$. Why are invariant polynomials of the roots rational, can you calculate them?

Question

I have been studying systems of equations based upon iterating a polynomial. The complete Galois group of these systems is only a subgroup of the permutation group. There are many invariant polynomials of the roots. My question is what are these invariants (rational, polynomial in c) and how can we calculate them? At this time I can calculate them numerically, but I cannot explain the intriguing results.
I looked at 3 systems, the first system, system 1, is based upon the roots of $P(P(P(x)))=x$, with $P(x)=X^2+c$. Discard the two solutions of $P(x)=x$ and we are left with 6 solutions which fall into 2 3 cycles:

$$x_{1 2}=P(x_{1 1})\quad; \quad x_{2 2}=P(x_{2 1})$$ $$x_{1 3}=P(x_{1 2})\quad; \quad x_{2 3}=P(x_{2 2})$$ $$x_{1 1}=P(x_{1 3})\quad; \quad x_{2 1}=P(x_{2 3})$$ The symmetry of these equations (Galois Group), $G_3$ is approximately $P_2 \times Z_3 \times Z_3$. The $P_2$ comes from interchanging the 2 cycles; $$x_{1 1} \leftrightarrow x_{2 1}$$ $$x_{1 2} \leftrightarrow x_{2 2}$$ $$x_{1 3} \leftrightarrow x_{2 3}$$ The $Z_3$'s come from the two cycles; $$x_{1 1} \rightarrow x_{1 2} \rightarrow x_{1 3} \rightarrow x_{1 1}$$ And similarly for cycle 2. System 2 used the same polynomial, but looked at the 5-cycles. $P(P(P(P(P(x)))))=x$, has 6 5-cycles. Unlike the previous case, which is solvable, this is not. Denote it's Galois group as $G_5$. It is very analogous to system 1, and I will skip some details. System 3 was based upon a cubic, $P(x)=x^3+a\,x+b$. System 3 looked at the 8 3-cycles. There is an infinite set of polynomials in the x's invariant under $G_3$ or $G_5$. I calculated some of these invariants, and found them to be low order polynomials in c. I will give several examples later. My question has 3 levels:

1. Are the invariants are rational in a, b or c. and why?
2. Are the invariants polynomial in a, b or c, and do they always have integer coefficients?
3. Brute force numerical calculation is not a very elegant way to calculate the invariants. Is there another way?

In system 1 the invariants were: $$S1=(x_{1 1}+x_{1 2}+x_{1 3})(x_{2 1}+x_{2 2}+x_{2 3})$$ $$S2=x_{1 1} x_{1 2}+x_{1 2} x_{1 3}+x_{1 3} x_{1 1}+x_{2 1} x_{2 2}+x_{2 2} x_{2 3}+x_{2 3} x_{2 1}$$ $$S3=x_{1 1} x_{1 2}^2+x_{1 2} x_{1 3}^2+x_{1 3} x_{1 1}^2+x_{2 1} x_{2 2}^2+x_{2 2} x_{2 3}^2+x_{2 3} x_{2 1}^2$$ $$S3r=x_{1 1}^2 x_{1 2}+x_{1 2}^2 x_{1 3}+x_{1 3}^2 x_{1 1}+x_{2 1}^2 x_{2 2}+x_{2 2}^2 x_{2 3}+x_{2 3}^2 x_{2 1}$$ $$S5=x_{1 1}^2 x_{1 2}^3+x_{1 2}^2 x_{1 3}^3+x_{1 3}^2 x_{1 1}^3+x_{2 1}^2 x_{2 2}^3+x_{2 2}^2 x_{2 3}^3+x_{2 3}^2 x_{2 1}^3$$

For system 2, the invariants were:

Note: $x_{i j}$ is normally defined for $j=1,2,...5$, but for these equations put $x_{i 6}=x_{i,1}$ $$T1=\Sigma_i(\Sigma_j x_{i j})^2$$ $$T2=\Sigma_i \Sigma_j x_{i j}x_{i (j+1)}$$ $$T3=\Sigma_i \Sigma_j x_{i j}x_{i (j+1)}^2$$ $$T4=\Pi_i(\Sigma_j x_{i j})$$

In system 3 the invariants were: $$R1=\Sigma_{i=1,8}(\Sigma_{j=1,3} X_{i j})^2$$ $$R2=\Sigma_{i=1,8}\Pi_{j=1,3}x_{i j}$$ $$R3=\Pi_{i=1,8}\Sigma_{j=1,3}x_{i j}$$

I calculated these invariants and found that they could be fit exactly by low order polynomials in c, as follows. $$S1=c+2$$ $$S2=2 c - 1$$ $$S3=3 c - 1$$ $$S3r= -5 c - 1$$ $$S5=3 c^2- 3 c - 1$$ $$T1=-22 c -5$$ $$T2=2 c - 1$$ $$T3=3 c - 1$$ $$T4=9 c^3+40 c^2+28 c + 32$$ $$R1=-12 a - 6 $$ $$R2 = b$$ $$R3=9 a^2+9 a + 9$$

As can be seen, the polynomials in c are about 1/2 the order of the polynomials in the x's and the coefficients are integers. A numerical calculation is not a proof and does not give much enlightenment.

Answer 1

I illustrate with system 1.

Define $P_6(x) = (P(P(P(x)))-x)/(P(x)-x)$ The polynomial division leaves no remainder.
The roots of $P_6(x)=0$ are $x_{1 1}, x_{1 2}, ... x_{2 3}$. From the coefficients of $P_6(x)$ one can calculate the sums of the powers over all the roots of $P_6(x)=0;

$$ \sigma_1 = \Sigma x_{i j}\\ \sigma_2 = \Sigma x_{i j}^2\\ etc. $$ These sums of powers are all polynomials in the coefficients of $P_6(x)$.

For the equations define $x_{i 4}= x_{i 1}$

$$ S2 = \Sigma_{i=1,2}\Sigma_{j=1,...3} x_{i j} x_{i j+1} $$

$$ x_{i j+1} = P(x_{i j)} $$ $$ S2=\Sigma x_{i j} P(x_{i j}) = \Sigma (x_{i j}^3+ c x_{i j}) = \sigma_3 + c \sigma_1 $$ The sums are over all solutions of $P_6(x)=0$.

S3 is obtained by $\Sigma x_{i j} P(x_{i j})^2$ and S3r is from $\Sigma x_{i j}^2 P(x_{i j})$.

For these and other invariants I verified the numerical results.

The algebra was quite complex, I used Mathematica. It looks like the $\sigma$'s are integers when the c is an integer, but I haven't proved it.

The general method seems to work for other polynomials and larger cycles