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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 '13 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 '13 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 '13 at 10:36
Ah, I missed the part about being simple. – John Wiltshire-Gordon Jul 26 '13 at 14:27
up vote 19 down vote accepted

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 '13 at 8:12
Thanks, Noam, that's great! – Jeremy Rickard Jul 26 '13 at 10:30
@Nick: I'm guessing he only checked words with exponents at most 3. – Jeremy Rickard Jul 26 '13 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 '13 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 '13 at 15:05

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 '13 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 '13 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 '13 at 2:19
@NoamD.Elkies It's funny that I got $|G|$ considering my 1st proof is wrong. – john mangual Jul 26 '13 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 '13 at 7:27

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