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I recently read the elegant paper Generalized Frobenius Schur Numbers, by Bump and Ginzburg, which I learned about here. The results in this paper imply the following:

Let $G$ be a finite group where every element is conjugate to its inverse. Then, for any $x \in G$, the number of ways to write $x$ as $ghg^{-1} h^{-1}$ is equal to the number of ways to write $x$ as $g^2 h^2$.

I'll give the proof below, using character theory. I tried to find an elementary proof and failed. Can you do it?


$\def\CC{\mathbb{C}}$We need to prove the equality $\sum_{g, h \in G} g^2 h^2 = \sum_{g, h \in G} g h g^{-1} h^{-1}$ in $\CC[G]$. Since we have $\CC[G] \cong \bigoplus_V \mathrm{End}(V)$, where the sum is over irreps of $V$, it is enough to show that both group elements act the same way on each $V$.

As I computed in this answer, $\sum_{g, h \in G} ghg^{-1} h^{-1}$ acts on an irrep $V$ by $\left( \frac{|G|}{\dim V} \right)^2$.

Now consider $\sum_{g \in G} g^2$ acting on an irrep $V$. Since $\sum_{g \in G} g^2$ is central in $\CC[G]$, it must act by a scalar, and computing traces we find that this scalar is $\frac{1}{\dim V} \sum_{g \in G} \chi_V(g^2)$. The quantity $\frac{1}{|G|} \sum_{g \in G} \chi_V(g)^2$ is known as the Frobenius-Schur indicator and is $\pm 1$ for a group where every element is conjugate to its inverse. So $\sum_{g \in G} g^2$ acts on $V$ by $\pm \frac{|G|}{dim V}$ and $\left( \sum_{g \in G} g^2 \right)^2$ acts on $V$ by $\left( \frac{|G|}{\dim V} \right)^2$.

We have shown that $\sum_{g, h \in G} ghg^{-1} h^{-1}$ and $\left( \sum_{g \in G} g^2 \right)^2 = \sum_{g,h \in G} g^2 h^2$ act on every irrep by the same scalar, so they are equal in the group algebra. QED.


The same argument shows that, in any finite group, the number of ways to write $x$ as $f^2 g^2 h^2$ is the same as the number of ways to write $x$ as $f^2 g h g^{-1} h^{-1}$, using that the Frob. Schur indicator is always $-1$, $0$ or $1$ and thus obeys $\epsilon^3 = \epsilon$. In case anyone finds that easier to think about...

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Partial result: if $g' = aga'$ and $h' = bhb'$ (where ' is inverse), then $$ghg'h' = ghaga'bhb' = (gha)^2 (gha)'ga'bhb' = (gha)^2 a'h'a'bhb' = (gha)^2(a'bhb'a'bhb') = (gha)^2(a'bhb')^2$$ –  Hurkyl Apr 18 '13 at 14:35
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2 Answers 2

up vote 5 down vote accepted

For $g \in G$, let $y_g$ be an element of $G$ such that $y_g g y_g^{-1} = g^{-1}$. For $g, h \in G$, let $x_{g,h} = gy_h^{-1}$.

Note that $ghg^{-1}h^{-1} = (gy_h^{-1}y_h)h(gy_h^{-1}y_h)^{-1}h^{-1} = xh^{-1}x^{-1}h^{-1}$ where $x = x_{g, h}$.

Then, $$ \displaystyle\sum_{g,h} ghg^{-1}h^{-1} = \displaystyle\sum_h \displaystyle\sum_g ghg^{-1}h^{-1} \\= \displaystyle\sum_h \displaystyle\sum_g (gy_h^{-1})h^{-1}(gy_h^{-1})^{-1}h^{-1} = \displaystyle\sum_h \displaystyle\sum_g gh^{-1}g^{-1}h^{-1} = \displaystyle\sum_{g, h}gh^{-1}g^{-1}h^{-1} $$

We also have $$\displaystyle\sum_{g,h} g^2h^2 = \displaystyle\sum_g \displaystyle\sum_h g^2h^2 \\ = \displaystyle\sum_g \displaystyle\sum_h g^2(g^{-1}h)^2 = \displaystyle\sum_g \displaystyle\sum_h ghg^{-1}h = \displaystyle\sum_{g, h} ghg^{-1}h = \displaystyle\sum_{g, h} gh^{-1}g^{-1}h^{-1}$$

For any $x \in G$, $(g, h) \mapsto (gy_h, y_h^{-1}g^{-1}h^{-1})$ is an explicit bijection from $\{(g, h) \in G \times G \mid ghg^{-1}h^{-1} = x\} $ to $\{(g, h)\in G \times G \mid g^{2}h^{2} = x\}$ with inverse $(g, h) \mapsto (gy_{h^{-1}g^{-1}}^{-1}, h^{-1}g^{-1})$.

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I think you could also calculate those directly, using this Number of ways a group element of a finite group can be written as a given word and the hypothesis that every element is conjugate to it's inverse to get the result. It all comes about calculating that $\sum_{(g,h) \in G^2} \chi(g^2h^2) = \frac{|G|^2}{\chi(1)}$ for all irreducible characters $\chi$, from which the result would follow (the calculations for the commutator case are fairly simple and yield the number of ways to write $x$ to be $|G| \sum_{\chi} \frac{\chi(x)}{\chi(1)}$ where the sum is taken over all irreducible characters).

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