Word evaluating to a group element and its inverse with different frequency

I'm supervising an undergraduate research project. Among other things, I've got the student to look at this paper of Gene Kopp and John Wiltshire-Gordon. This question arose from a missing complex conjugate in something the student wrote.

Let $g$ be an element of a finite group $G$, and $w$ a word in $n$ variables. If you evaluate the word on all $n$-tuples of elements of $G$, does it give $g$ and $g^{-1}$ the same number of times?

I thought the answer must be "no", but found it frustratingly difficult to come up with an example. After tapping local knowledge it seems that I was right. There's a recent paper of Alexander Lubotzky that proves that if $G$ is a finite simple group then the only restriction on a subset $A\subseteq G$ for it to be the image of the word map for some word in 2 variables is that it contains the identity and is fixed by $\operatorname{Aut}(G)$. Since there are finite simple groups (e.g., the Mathieu group $M_{11}$) with elements that are not sent to their inverses by any automorphism, this answers the question.

However, my real question is whether there's a relatively simple example?

Lubotzky's paper doesn't give an explicit word, although it does show that, for $M_{11}$, there's a word that works with length at most about $1.7\times 10^{244552995}$.

Presumably one can do a bit better than that?

There are obvious restrictions on $g\in G$ and $w$ that rule out really small examples. There can't be any automorphism of $G$ sending $g$ to $g^{-1}$ or any automorphism of the free group $F_n$ sending $w$ to $w^{-1}$.

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GAP's SmallGroup(64,28) has elements not automorphic to their inverses. Some experimentation suggests that the tricky bit is finding the word. –  John Wiltshire-Gordon Jul 26 at 1:32
@John: Indeed. And the existence of such an element doesn't guarantee the existence of a word that works. For example, in the Frobenius group of order 20, elements of order 4 are not automorphic to their inverses, but it's not hard to show that no word works there. –  Jeremy Rickard Jul 26 at 6:42
I think SmallGroup(64,28) can't give an example for much the same reason as the Frobenius group of order 20. For the elements $g$ not automorphic to their inverses there are quotient maps $\pi:G\to H$ so that the conjugacy class of $g$ is $\pi^{-1}(h)$ for some $h\in H$, which reduces the question to the smaller group $H$. Maybe there's a theorem about soluble groups here somewhere? –  Jeremy Rickard Jul 26 at 10:36
Ah, I missed the part about being simple. –  John Wiltshire-Gordon Jul 26 at 14:27

Yes, one can do much better than $1.7 \times 10^{244552995}$ (not surprisingly, because we're asking less than Lubotzky: one of the two counts must be less than the other, but not necessarily zero). In fact a word of length $10$ suffices.

I tried $G = M_{11}$ and $g$ an element of order $11$, and took $n=2$, which makes exhaustive computation easily feasible (the first variable can be assumed to lie in one of the $10$ conjugacy classes so there's only $10 \, |M_{11}| = 79200$ group elements to compute given $w$). None of the words $w(x,y) = x^a y^b x^c y^d$ seems to work, but several of the form $w(x,y) = x^a y^b x^c y^d x^e$ solve the problem. The first one (in lexicographic order) with all exponents at most $3$ is $(a,b,c,d,e) = (1,2,1,3,3)$, i.e. $w(x,y) = x y^2 x y^3 x^3$, for which $w(x,y) = g$ has $7491$ solutions but $w(x,y) = g^{-1}$ has only $7458$.

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This is a great answer and very intriguing, but I'm slightly confused. Surely one can conjugate $w(x,y)=x^a y^b x^c y^d x^e$ by $x^{-e}$ to get a word with only four powers... Which seems to contradict your suggestion that four powers isn't enough, but five is. What am I missing? –  Nick Gill Jul 26 at 8:12
Thanks, Noam, that's great! –  Jeremy Rickard Jul 26 at 10:30
@Nick: I'm guessing he only checked words with exponents at most 3. –  Jeremy Rickard Jul 26 at 10:31
Thanks $-$ and yes, $x^4 y^2 x y^3$ works as @Nick Gill explained (I just re-checked this directly), so I'll change to this simpler formula in the next edit. –  Noam D. Elkies Jul 26 at 13:23
By the way, this teaches me a lesson about laziness. I did test several words on $M_{11}$. But when I realized that it took to long to loop over all pairs of elements, I "saved" myself a few minutes typing by looping over a few thousand random pairs of elements, rather than doing it properly, hoping to find an example with an obvious discrepancy that I could look at more closely. But I was never going to spot an example where the counts were as close as 7491 and 7458. –  Jeremy Rickard Jul 26 at 15:05
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Check out Corollary 1 in Winter, David L. The automorphism group of an extraspecial p-group. Rocky Mountain J. Math. 2 (1972), no. 2, 159–168. 20B25

It gives the example you need.

EDIT A further search reveals this (not so old) MO discussion: element algebraically distinguishable from its inverse

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Sorry, why? This gives you the group $G$, but what is the word $w$? –  David Speyer Jul 26 at 1:21
@DavidSpeyer Your point is well-taken, but I was on the tangent of finding a group where some $g$ and $g^{-1}$ are not in an automorphism orbit, so this answers a part of the OP's question. –  Igor Rivin Jul 26 at 1:29

They are solving for $\gamma_G(w) = \# \{ t \in G^n : w(t)=1 \}$. For words that define surfaces they get a count in terms of the characters of $G$: $$\gamma_G(w)= \big|G\big|^{n-1} \underbrace{\sum_{\rho \in \mathrm{Irr}(G)} (\dim \rho)^k \langle \rho|g\rangle}_{\zeta_G(-k)}$$

This formula appears in many places, e.g. arXiv:0905.0731:Topological Quantum Field Theories from Compact Lie Groups. They are using the fact the characters of a group form a TQFT.

I don't really understand why people don't study group statistics using this type of result. You can generalize the bound $\gamma_G(aba^{-1}b^{-1})\leq 5/8$ easily.

Have you tried a word like $a^2 b$ ? I think this proves that $a^2 b$ does not work:

$$\gamma_G(a^2 bc)= \big| \big\{ a,b: a^2bc = 1 \big\}\big|= \big| \big\{ a,b: abca^{-1} = 1 \big\}\big| = |G| \cdot\big| \big\{ b: bc = 1 \big\}\big| = |G|$$ but then $\gamma_G(a^2 bc) = \gamma_G(a^2 bc^{-1})$.

More succinctly, $(a^2b)^{-1} = b^{-1}a^{-2}$ however this is induced by automorphism of the free group: $$a^2 b \mapsto b^{-1}a^2 \mapsto b^{-1}a^{-2}$$

Also, as you mentioned $g, g^{-1}$ should not be conjugate (e.g. alternating groups have an outer automorphism, $Aut(A_n) = S_n$).

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More succinctly yet, you can solve $a^2 b = g$ for $b$, so every $g$ arises $|G|$ times... –  Noam D. Elkies Jul 26 at 2:19
@NoamD.Elkies It's funny that I got $|G|$ considering my 1st proof is wrong. –  john mangual Jul 26 at 2:34
In fact, any word of the form $a^ib^ja^k$ is sent to its inverse by an automorphism of $F_2$, so any candidate word must be more complicated than that. –  Jeremy Rickard Jul 26 at 7:27