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You were probably mainly interested in the case of $a\le n$; the other case of $a>n$ seems to be false. Here's a quick counterexample for $a>n$, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

If there were a factorization with $a\le n$, it would need to involve nonsymmetric polynomials. But since $S$ is a unique factorization domain, this would mean for any nonsymmetric factor, we would also need all the elements of its orbit as factors. While I was editing this, Will Sawin came up with an elegant proof, whereas I was just grinding through cases, so I will omit those cases.

Since Will mentions still needing the base case of $n=a=3$, notice that we would need a nonsymmetric factor, which means a nonsymmetric factor with either (a) terms of the form $x_ix_j^2$, with nontrivial orbit, yielding too high a degree, or (b) it has terms of the form $x_i$ with some included and some omitted, contradicting the product including all $x_i^3$ terms, or (c) it has some terms $x_ix_j$ with $i\ne j$ and omits others, again necessitating other factors in the orbit, giving too high a degree for the product, or (d) it has some terms $x_i^2$ and not others, giving the same contradiction.

You were probably mainly interested in the case of $a\le n$. Here's a quick counterexample for $a>n$, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

If there were a factorization with $a\le n$, it would need to involve nonsymmetric polynomials. But since $S$ is a unique factorization domain, this would mean for any nonsymmetric factor, we would also need all the elements of its orbit as factors. While I was editing this, Will Sawin came up with an elegant proof, whereas I was just grinding through cases, so I will omit those cases.

Since Will mentions still needing the base case of $n=a=3$, notice that we would need a nonsymmetric factor, which means a nonsymmetric factor with either (a) terms of the form $x_ix_j^2$, with nontrivial orbit, yielding too high a degree, or (b) it has terms of the form $x_i$ with some included and some omitted, contradicting the product including all $x_i^3$ terms, or (c) it has some terms $x_ix_j$ with $i\ne j$ and omits others, again necessitating other factors in the orbit, giving too high a degree for the product, or (d) it has some terms $x_i^2$ and not others, giving the same contradiction.

This is true for $a\le n$, which is probably the case you were mainly interested in, but false for $a>n$. Here's a quick counterexample for $a>n$ first, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

Now to a proof that you are right for $a\le n$. The fact that $S$ is a unique factorization domain seems to imply that if there were a factorization of some $h_a$ with $a\le n$, it would either have to be into symmetric polynomials, or for each nonsymmetric factor one would need all elements of its $S_n$-orbit as well. For $a \le n$, one cannot factor $h_a$ into a nontrivial product of symmetric polynomials.

If there were a factorization into irreducibles, not all of which are symmetric polynomials, then each irreducible would still need to have one or more summands of the form $x_i^d$, in order for the product of the irreducibles to have a summand of the form $x_i^a$. Now the size of the orbit of the irreducible would be at least ${a \choose |S|}$, where $S$ is the set of indices $i$ so that $x_i^d$ appears as a summand in our factor. But ${a\choose |S|} \ge a$, provided our factor is nonsymmetric, giving a lower bound of $a$ on the number of irreducibles in the product, and hence on the degree of the product of the irreducibles, where this bound can only be achieved if all of the irreducibles are homogeneous of degree 1. It is easy to check this cannot happen. Maybe someone else will have a more elegant proof. (I apologize for the repeated edits.)

You were probably mainly interested in the case of $a\le n$; the other case of $a>n$ seems to be false. Here's a quick counterexample for $a>n$, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

If there were a factorization with $a\le n$, it would need to involve nonsymmetric polynomials. But since $S$ is a unique factorization domain, this would mean for any nonsymmetric factor, we would also need all the elements of its orbit as factors. While I was editing this, Will Sawin came up with an elegant proof, whereas I was just grinding through cases, so I will omit those cases.

Since Will mentions still needing the base case of $n=a=3$, notice that we would need a nonsymmetric factor, which means a nonsymmetric factor with either (a) terms of the form $x_ix_j^2$, with nontrivial orbit, yielding too high a degree, or (b) it has terms of the form $x_i$ with some included and some omitted, contradicting the product including all $x_i^3$ terms, or (c) it has some terms $x_ix_j$ with $i\ne j$ and omits others, again necessitating other factors in the orbit, giving too high a degree for the product, or (d) it has some terms $x_i^2$ and not others, giving the same contradiction.

This is true for $a\le n$, which is probably the case you were mainly interested in, but false for $a>n$. Here's a quick counterexample for $a>n$ first, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

Now to a proof that you are right for $a\le n$. The fact that $S$ is a unique factorization domain seems to imply that if there were a factorization of some $h_a$ with $a\le n$, it would either have to be into symmetric polynomials, or for each nonsymmetric factor one would need all elements of its $S_n$-orbit as well. For $a \le n$, one cannot factor $h_a$ into a nontrivial product of symmetric polynomials.

If there were a factorization into irreducibles, not all of which are symmetric polynomials, then each irreducible would still need to have one or more summands of the form $x_i^d$, in order for the product of the irreducibles to have a summand of the form $x_i^a$. Now the size of the orbit of the irreducible would be at least ${a \choose |S|}$, provided our factor is nonsymmetric, where $S$ is the set of indices $i$ so that $x_i^d$ appears as a summand in our factor. But ${a\choose |S|} \ge a$, giving a lower bound of $a$ on the number of irreducibles in the product, and hence on the degree of the product of the irreducibles, where this bound can only be achieved if all of the irreducibles are homogeneous of degree 1. It is easy to check this cannot happen. Maybe someone else will have a more elegant proof.

This is true for $a\le n$, which is probably the case you were mainly interested in, but false for $a>n$. Here's a quick counterexample for $a>n$ first, letting $n=1$ and $a=2$. Notice then that $h_2 = (h_1)^2$.

Now to a proof that you are right for $a\le n$. The fact that $S$ is a unique factorization domain seems to imply that if there were a factorization of some $h_a$ with $a\le n$, it would either have to be into symmetric polynomials, or for each nonsymmetric factor one would need all elements of its $S_n$-orbit as well. For $a \le n$, one cannot factor $h_a$ into a nontrivial product of symmetric polynomials.

If there were a factorization into irreducibles, not all of which are symmetric polynomials, then each irreducible would still need to have one or more summands of the form $x_i^d$, in order for the product of the irreducibles to have a summand of the form $x_i^a$. Now the size of the orbit of the irreducible would be at least ${a \choose |S|}$, where $S$ is the set of indices $i$ so that $x_i^d$ appears as a summand in our factor. But ${a\choose |S|} \ge a$, provided our factor is nonsymmetric, giving a lower bound of $a$ on the number of irreducibles in the product, and hence on the degree of the product of the irreducibles, where this bound can only be achieved if all of the irreducibles are homogeneous of degree 1. It is easy to check this cannot happen. Maybe someone else will have a more elegant proof. (I apologize for the repeated edits.)