The degree of a constant polynomial on a finite group

Question

A function $f:X\to X$ on a group $X$ is called a polynomial if there exists $n\in\mathbb N=\{1,2,\dots\}$ and elements $a_0,a_1,\dots,a_n\in X$ such that $f(x)=a_0xa_1x\cdots xa_n$ for all $x\in X$. The smallest number $n\in\mathbb N$ in such a representation is called the degree of the polynomial $X$.

The constant function $1_X:X\to\{1\}\subseteq X$ on a finite group $X$ is a polynomial of degree $\le\exp(X):=\min\{n\in\mathbb N:\forall x\in X\;\;(x^n=1)\}$ because $f(x)=1=x^{\exp(X)}$ for any $x\in X$.

Each polynomial $f$ on a commutative group $X$ is of form $f(x)=ax^n$ for some $a\in X$ and $n\le\exp(X)$, which implies that $\deg(1_X)=\exp(X)$ for any finite commutative group $X$.

Problem 1. Given a non-commutative finite group $X$, calculate (or evaluate) the degree $\deg(1_X)$ of the constant polynomial $1_X:X\to\{1\}\subseteq X$.

Remark 1. Calculations in GAP show that for any non-commutative group $X$ of order $|X|<20$, except for $A_4$ and $C_3\times S_3$, the degree of the constant function $1_X$ is equal $4$. For the groups $A_4$ and $C_3\times S_3$ this degree equals $6$.

Remark 2. It can be shown that $\deg(1_X)=4$ for any dihedral group $X=D_{2n}$ (because for any non-central element $b$ of order $2$ in $D_{2n}$ we have $bxxbxx=1$ for all $x\in D_{2n}$).

Remark 3. It is easy to see that $\deg(1_X)\ge \exp(Z(X))$, where $Z(X)=\{z\in X:\forall x\in X\;\;(xz=zx)\}$ is the center of $X$.

Keith Kearnes in his comment observed that $\deg(1_X)\ge\deg(1_{X/N})\ge\exp(Z(X/N))$ for any normal subgroup $N$ of $X$. In particular, $\deg(1_X)\ge \exp(X/[X,X])$.

Problem 2. What can be said about the degree of the constant polynomial $1_X$ on a finite simple group $X$. Is $\deg(1_X)=\exp(X)$? Is this equality true for the simple group $X=A_5$?

Remark 4. It seems that this problem has been considered by various authors in the contexts of strong laws on groups. In particular, by Corollary 1 in this paper of Schneider and Thom, for the symmetric group $X=S_n$ we have $\deg(1_X)\ge \frac{n}4$, which implies that the alternating group $X=A_n$ has $\deg(1_X)\ge\frac{n}8$.

Answer 1

If $X = A_5$ then $\exp(X) = \mathop{\rm lcm}(2,3,5) = 30$ but $\deg 1_X \leq 10$. Indeed if $c \in X$ is a 5-cycle then $$ f(x) := x c x c^2 x c^3 x c^4 x $$ satisfies $f(x)^2 = 1$ for all $x \in X$ (in fact $f(x)$ is a double transposition unless $x$ is in the subgroup generated by $c$ in which case of course $f(x)=1$).

added later:

To answer questions in the comments: I don't have a structural explanation why $f$ works, and a similar construction barely fails in $S_5$ (see below) and doesn't seem to work at all in $A_6$, $A_7$ and beyond.

This $f$ was found computationally by constructing the $A_5$ multiplication table and searching for "polynomials" whose image misses some exponents. Here's what the same technique finds for some other small groups:

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$X = S_4$ (exponent $12$): $\deg 1_X = 6$, attained for example by $(x c x c^2 x)^2$ where $c$ is a 3-cycle. If I computed correctly, all degree-6 polynomials that represent $1_{S_4}$ are squares.

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$X = S_5$ (exponent $60$): $\deg 1_X \leq 20$. This follows from $\deg 1_{A_5} \leq 10$ and the inequality $$ \deg 1_X \leq \deg 1_N \cdot \deg 1_{X/N} $$ for any normal subgroup $N$ of $X$. To prove this inequality, let $g$ be a polynomial on $X/N$ that represents the identity, and lift it arbitrarily to a polynomial $\tilde g$ on $X$. Then $\tilde g$ maps $X$ to $N$, so $\phi \circ \tilde g = 1_X$ for any polynomial $\phi$ on $N$ that represents $1_N$, and $\deg(\phi \circ \tilde g) = (\deg \phi) (\deg \tilde g) = (\deg \phi) (\deg g)$.

Taking $N = A_5$ this yields the degree-20 polynomial $f(x^2)^2$ where $f(x) = x c x c^2 x c^3 x c^4 x$ as before. In the comments Taras Banakh asked whether $f^2$ itself would work; curiously it almost does: for an odd permutation $x$, the exponent of $f(x)$ is still $2$ $-$ except when $x$ is one of the 10 $4$-cycles in the normalizer of $\langle c \rangle$, when $f(x)$ has exponent $4$. This does mean that $f(x)^4$ is another degree-$20$ representation of $1_{S_5}$.

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$X = {\rm GL}_3({\bf Z}/2{\bf Z})$ (the second-smallest noncyclic simple group, which has order $168$ and exponent $84)$: $\deg 1_X \leq 36$, attained by $(xbxcx)^{12}$ where $b$ and $c$ have order $2$ and $bc$ has order $4$. (It turns out that for such $b,c$ no value of $xbxcx$ has exponent $7$, though each of the remaining exponents $1,2,3,4$ does occur.)

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Probably the upper bounds for $S_5$ and ${\rm GL_3}({\bf Z}/2{\bf Z})$ are not optimal; I wouldn't even be too surprised if $\deg 1_{A_5}$ is smaller than $10$.