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A finite group with more than 3/4 of elements of order 2 is abelian

Let \(G\) be a group (not necessarily finite) and assume that each element of \(G\) has order 2, that is, \(g^2=1\) for all \(g\in G\) (in multiplicative notation). Then \(G\) is abelian. Indeed, we have that \(g^{-1}=g\) for all \(g\in G\) and hence

\[[x,y]=x^{-1}y^{-1}xy=(xy)(xy)=(xy)^2=1\]for all \(x,y\in G\), and so \(xy=yx\) for all \(x,y\in G\).

The previous (very elementary) observation motivates the following guess: if \(G\) is finite and if 'most' of its elements has order 2 then \(G\) must be abelian. This guess reveals to be true and its proof is quite funny.

**Theorem**. *Let \(G\) be a finite group and let \[\Omega=\{g\in G : g^2=1\}.\]If \(|\Omega|>\frac{3}{4}|G|\), then \(G\) is abelian.*

Proof. Let \(x\in\Omega\). Then \[|x\Omega\cap\Omega|=|x\Omega|+|\Omega|-|x\Omega\cup\Omega|>\frac{3}{2}|G|-|G|=\frac{1}{2}|G|.\]Now let \(y\in x\Omega\cap\Omega\). Then \(xy\in x(x \Omega )=\Omega\), so \(x,y,xy\in\Omega\). Therefore \([x,y]=1\) as above, and so \(y\in C_G(x)\), the centralizer of \(x\) in \(G\). Thus \(x\Omega\cap\Omega\subseteq C_G(x)\) and so \(|C_G(x)|>\frac{1}{2} |G|\). By Lagrange's theorem we must have \(C_G(x)=G\), so \(x\in Z(G)\) (the center of the group). Hence \(\Omega\subseteq Z(G)\) and so \(|Z(G)|>\frac{3}{4}|G|\). Again by Lagrange's theorem we conclude that (Z(G)=G\), so \(G\) is abelian.

We note that the bound \(|\Omega|>\frac{3}{4}|G|\) cannot be improved. Indeed, the dihedral group \[D_4=\langle a,b|a^4=b^2=(ab)^2=1\rangle\]has 8 elements and 6 elements of order 2, but \(D_4\) is not abelian.

\[[x,y]=x^{-1}y^{-1}xy=(xy)(xy)=(xy)^2=1\]for all \(x,y\in G\), and so \(xy=yx\) for all \(x,y\in G\).

The previous (very elementary) observation motivates the following guess: if \(G\) is finite and if 'most' of its elements has order 2 then \(G\) must be abelian. This guess reveals to be true and its proof is quite funny.

Proof. Let \(x\in\Omega\). Then \[|x\Omega\cap\Omega|=|x\Omega|+|\Omega|-|x\Omega\cup\Omega|>\frac{3}{2}|G|-|G|=\frac{1}{2}|G|.\]Now let \(y\in x\Omega\cap\Omega\). Then \(xy\in x(x \Omega )=\Omega\), so \(x,y,xy\in\Omega\). Therefore \([x,y]=1\) as above, and so \(y\in C_G(x)\), the centralizer of \(x\) in \(G\). Thus \(x\Omega\cap\Omega\subseteq C_G(x)\) and so \(|C_G(x)|>\frac{1}{2} |G|\). By Lagrange's theorem we must have \(C_G(x)=G\), so \(x\in Z(G)\) (the center of the group). Hence \(\Omega\subseteq Z(G)\) and so \(|Z(G)|>\frac{3}{4}|G|\). Again by Lagrange's theorem we conclude that (Z(G)=G\), so \(G\) is abelian.

We note that the bound \(|\Omega|>\frac{3}{4}|G|\) cannot be improved. Indeed, the dihedral group \[D_4=\langle a,b|a^4=b^2=(ab)^2=1\rangle\]has 8 elements and 6 elements of order 2, but \(D_4\) is not abelian.

Lipschitz functions are a powerful and common tool in mathematical analysis, since they can be naturally defined in any metric space. For my purpose, I restrict my discussion to the Euclidean space \(\mathbb{R}\).

Recall that a function \(f\colon A\to\mathbb{R}^n\) is a Lipschitz function defined a non empty set \(A\subseteq\mathbb{R}\) if \(|f(x)-f(y)|\le L|x-y|\) for all \(x,y\in\mathbb{R}\). The optimal constant \(L\) is the Lipschitz constant of \(f\) and in this case \(f\) is also called \(L\)-Lipschitz.

Note that Lipschitz functions are (absolutely) continuous. Moreover, Lipschitz functions are well-behaved with respect to the property of extension and differentiation.

The following theorem on the extension property of Lipschitz functions is due to Mojżesz David Kirszbraun, who proved it in the more general context of Hilbert spaces.

**Theorem** (Kirszbraun). *Let \(A\subseteq\mathbb{R}^m\). For any \(L\)-Lipschitz function \(f\colon A\to\mathbb{R}^n\) there exists a \(L\)-Lipschitz function \(F\colon\mathbb{R}^m\to\mathbb{R}^n\) such that \(F=f\) on \(A\).*

The following theorem, instead, is the celebrated Hans Rademacher's theorem on differentiation of Lipschitz functions.

**Theorem** (Rademacher). *Let \(A\subseteq\mathbb{R}^m\) be an open set. If \(f\colon A\to\mathbb{R}^n\) is a Lipschitz function, then \(f\) is differentiable almost everywhere on \(A\).*

Another important feature of Lipschitz functions is their natural relation with the Hausdorff measure. Indeed, if \(f\colon\mathbb{R}^n\to\mathbb{R}^n\) is \(L\)-Lipschitz and \(E\) is a measurable set, then \(\mathcal{H}^s(f(E))\le L^s\mathcal{H}^s(E)\) for all \(0\le s\le n\), where \(\mathcal{H}^s\) denotes the \(s\)-dimensional Hausdorff measure.

This particular property motivates the following problem, due to Miklós Laczkovich.

**Problem** (Laczkovich). *Let \(n\geq1\) and let \(E\subseteq\mathbb{R}^n\) a (compact) measurable set with positive Lebesgue measure, \(|E|>0\). Is there a Lipschitz function \(f\colon\mathbb{R}^n\to\mathbb{R}^n\) such that \(f(E)=[0,1]^n\)?*

In other words, the problem asks whether a measurable set with positive measure can be 'compressed' into a full box by a Lipschitz function, and hence is properly a problem related to the geometric properties of measurable sets. Note that, by the observation above, the hypothesis that \(|E|>0\) is necessary. Moreover, note that it is enough to prove that \(f(E)\) has non empty interior, since the projection onto a ball is a \(1\)-Lipschitz map.

It is an exercise to see that the problem has positive answer when \(n=1\). Indeed, let \(f\colon\mathbb{R}\to\mathbb{R}\) the 'primitive' of the characteristic function \(1_E\) of the compact set \(E\), that is, for any \(x\in\mathbb{R}\) we let \[f(x)=|E\cap(-\infty,x)|=\int_{-\infty}^x 1_E(t)dt.\]The function \(f\) is well-defined on \(\mathbb{R}\) because \(E\) is compact. Moreover \(f\) is increasing and \(1\)-Lipschitz. Note that \(f\) is also constant on every connected component of \(\mathbb{R}\setminus E\), hence \(f(E)\) and \(f(\mathbb{R})\) coincide. But \(f(E)=[\min E,|E|]\), because \(f\) is non constant, since \(|E|>0\).

The problem has also positive answer when \(n=2\), but the solution is much harder and involves a completely different approach. The first result in this direction has been obtained by N. X. Uy in 1979.

**Theorem** (Uy). *For any compact set \(E\subseteq\mathbb{R}^2\) with positive Lebesgue measure, \(|E|>0\), there exists a non constant Lipschitz function with complex values \(f\colon\mathbb{R}^2\to\mathbb{C}\) which is holomorphic everywhere on the complement of \(E\) (infinity included).*

Hence, if we identify \(\mathbb{C}\) with \(\mathbb{R}^2\), this theorem proves the existence of a Lipschitz map \(f\colon\mathbb{R}^2\to\mathbb{R}^2\) that preserves the orientation and that is open on the complement of \(E\). Therefore, using the degree theory, it follows that \(f(\mathbb{R}^2\setminus E)\subseteq f(E)\) and that \(f(\mathbb{R}^2\setminus E)\) is open.

More recently, another approach has been developed. The proof does not involve complex analysis. The argument of the proof, developed for the first time by D. Preiss in 1992, has successively been improved by J. Matousek, in 1997. In 2004, G. Alberti, M. Csörnyei e D. Preiss have simplified the original argument of Matousek and have applied it to other results.

The core of the proof relies on the following combinatorical result, which can be seen as a geometric version of Dilworth's Theorem. Here, by \(x\)-curve (\(y\)-curve), we mean the graph of a \(1\)-Lipschitz function \(y=y(x)\) (\(x=x(y)\)) defined for all \(x\in\mathbb{R}\) (\(y\in\mathbb{R}\)).

**Lemma**. *Any set \(S\subset\mathbb{R}^2\) of \(m\) points can be covered using at most \(\sqrt{m}\) \(x\)-curves and \(\sqrt{m}\) \(y\)-curves.*

When \(n\ge3\), the answer to the Laczkovich problem is completely open. In particular, the combinatorical argument contained in the Lemma above is not strong enough to provide a solution similar to the case \(n=2\).

To conclude, we note that a possible way to proceed in order to solve the problem for \(n\ge2\) (similarly to the case \(n=1\)), is to solve the the equation \[\text{det}(Df) = 1_E,\]where \(Df\) is the Jacobian matrix of \(f\), on some bounded smooth domain \(\Omega\) in \(\mathbb{R}^n\) contanining \(E\) imposing a Dirichlet boundary condition in such a way that \(f(\Omega)\) has non empty interior. If such a solution \(f\) exists, then \(f(\Omega\setminus E)\) has null measure, and thus \(f(E)\) coincide with \(f(\Omega)\). The difficulty of this approach is that in general the equation \[\text{det}(Df) = g,\]known as the*Jacobian equation*, does not admit any solution \(f\) that is a Lipschitz omeomorphism even if the datum \(g\) is continuous and positive. Similar considerations hold for the equation \[\text{div} f = g,\]known as the *divergence equation*.

Recall that a function \(f\colon A\to\mathbb{R}^n\) is a Lipschitz function defined a non empty set \(A\subseteq\mathbb{R}\) if \(|f(x)-f(y)|\le L|x-y|\) for all \(x,y\in\mathbb{R}\). The optimal constant \(L\) is the Lipschitz constant of \(f\) and in this case \(f\) is also called \(L\)-Lipschitz.

Note that Lipschitz functions are (absolutely) continuous. Moreover, Lipschitz functions are well-behaved with respect to the property of extension and differentiation.

The following theorem on the extension property of Lipschitz functions is due to Mojżesz David Kirszbraun, who proved it in the more general context of Hilbert spaces.

The following theorem, instead, is the celebrated Hans Rademacher's theorem on differentiation of Lipschitz functions.

Another important feature of Lipschitz functions is their natural relation with the Hausdorff measure. Indeed, if \(f\colon\mathbb{R}^n\to\mathbb{R}^n\) is \(L\)-Lipschitz and \(E\) is a measurable set, then \(\mathcal{H}^s(f(E))\le L^s\mathcal{H}^s(E)\) for all \(0\le s\le n\), where \(\mathcal{H}^s\) denotes the \(s\)-dimensional Hausdorff measure.

This particular property motivates the following problem, due to Miklós Laczkovich.

In other words, the problem asks whether a measurable set with positive measure can be 'compressed' into a full box by a Lipschitz function, and hence is properly a problem related to the geometric properties of measurable sets. Note that, by the observation above, the hypothesis that \(|E|>0\) is necessary. Moreover, note that it is enough to prove that \(f(E)\) has non empty interior, since the projection onto a ball is a \(1\)-Lipschitz map.

It is an exercise to see that the problem has positive answer when \(n=1\). Indeed, let \(f\colon\mathbb{R}\to\mathbb{R}\) the 'primitive' of the characteristic function \(1_E\) of the compact set \(E\), that is, for any \(x\in\mathbb{R}\) we let \[f(x)=|E\cap(-\infty,x)|=\int_{-\infty}^x 1_E(t)dt.\]The function \(f\) is well-defined on \(\mathbb{R}\) because \(E\) is compact. Moreover \(f\) is increasing and \(1\)-Lipschitz. Note that \(f\) is also constant on every connected component of \(\mathbb{R}\setminus E\), hence \(f(E)\) and \(f(\mathbb{R})\) coincide. But \(f(E)=[\min E,|E|]\), because \(f\) is non constant, since \(|E|>0\).

The problem has also positive answer when \(n=2\), but the solution is much harder and involves a completely different approach. The first result in this direction has been obtained by N. X. Uy in 1979.

Hence, if we identify \(\mathbb{C}\) with \(\mathbb{R}^2\), this theorem proves the existence of a Lipschitz map \(f\colon\mathbb{R}^2\to\mathbb{R}^2\) that preserves the orientation and that is open on the complement of \(E\). Therefore, using the degree theory, it follows that \(f(\mathbb{R}^2\setminus E)\subseteq f(E)\) and that \(f(\mathbb{R}^2\setminus E)\) is open.

More recently, another approach has been developed. The proof does not involve complex analysis. The argument of the proof, developed for the first time by D. Preiss in 1992, has successively been improved by J. Matousek, in 1997. In 2004, G. Alberti, M. Csörnyei e D. Preiss have simplified the original argument of Matousek and have applied it to other results.

The core of the proof relies on the following combinatorical result, which can be seen as a geometric version of Dilworth's Theorem. Here, by \(x\)-curve (\(y\)-curve), we mean the graph of a \(1\)-Lipschitz function \(y=y(x)\) (\(x=x(y)\)) defined for all \(x\in\mathbb{R}\) (\(y\in\mathbb{R}\)).

When \(n\ge3\), the answer to the Laczkovich problem is completely open. In particular, the combinatorical argument contained in the Lemma above is not strong enough to provide a solution similar to the case \(n=2\).

To conclude, we note that a possible way to proceed in order to solve the problem for \(n\ge2\) (similarly to the case \(n=1\)), is to solve the the equation \[\text{det}(Df) = 1_E,\]where \(Df\) is the Jacobian matrix of \(f\), on some bounded smooth domain \(\Omega\) in \(\mathbb{R}^n\) contanining \(E\) imposing a Dirichlet boundary condition in such a way that \(f(\Omega)\) has non empty interior. If such a solution \(f\) exists, then \(f(\Omega\setminus E)\) has null measure, and thus \(f(E)\) coincide with \(f(\Omega)\). The difficulty of this approach is that in general the equation \[\text{det}(Df) = g,\]known as the

G. Alberti, M. Csörnyei, and D. Preiss,

Mathematics. Proceedings of the 4th Congress (4ECM), 2005, pp. 3-22.

G. Alberti, M. Csörnyei, and D. Preiss,

D. Burago and B. Kleiner,

8 (1998), 273-282.

R. P. Dilworth,

J. Heinonen,

M. Laczkovich,

(1991), 145-176.

M. Laczkovich,

J. Matousek,

D. Preiss,

N. X. Uy,

I found this nice Italian crossword at Maddmaths! and I decided to rewrite it with LaTeX (using the cwpuzzle package). You can download it here. Have fun!

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