[Algebra] A nice group automorphism

[Paolo90]

Problem. Let \( \displaystyle {G} \) be a finite group and let \( \displaystyle {T}\in\text{Aut}{\left({G}\right)} \) (the group of the automorphisms of \( \displaystyle {G} \)). Suppose that \( \displaystyle {T} \) maps more than 3/4 of the elements of \( \displaystyle {G} \) to their inverse, i.e. \( \displaystyle {T}{x}={{x}}^{{-{1}}} \) for more than 3/4 of the elements of \( \displaystyle {G} \).

Prove that \( \displaystyle {G} \) is abelian and \( \displaystyle {T}{x}={{x}}^{{-{1}}} \), \( \displaystyle \forall{x}\in{G} \).

[dissonance]

Dear Paolo, please do not reveal the solution to this exercise! I'm willing to test myself against it in an attempt to revive my long-time dormant group theory skills. (But I'll need some time as I'm busy...).

[Paolo90]

Ok, dissonance.

Do not worry, take your time and enjoy group theory

[Paolo90]

Testo nascosto, fai click qui per vederlo

Proof. Let \( \displaystyle {Y}\:={\left\lbrace{g{\in}}{G}:{T}{g{=}}{{g}}^{{-{1}}}\right\rbrace} \). Then \( \displaystyle {\left|{Y}\right|}\gt{\frac{{{3}}}{{{4}}}}{\left|{G}\right|} \) by hypothesis. Choose \( \displaystyle {a}\in{Y} \) and set \( \displaystyle {a}{Y}=:{Z} \). It's easy to see that \( \displaystyle {\left|{Z}\right|}={\left|{Y}\right|} \). Now we want to compute \( \displaystyle {\left|{Y}\cap{Z}\right|} \). Since \( \displaystyle {\left|{G}\backslash{Y}\right|}={\left|{G}\backslash{Z}\right|}\le\frac{{1}}{{4}}{\left|{G}\right|} \), simply using the De Morgan laws, we get \( \displaystyle {\left|{Y}\cap{Z}\right|}\gt{\frac{{{\left|{G}\right|}}}{{{2}}}} \). Now pick \( \displaystyle {s}\in{Y}\cap{Z} \) and \( \displaystyle {y}\in{Y} \): \( \displaystyle {s} \) must be of the form \( \displaystyle {a}{y} \) and since \( \displaystyle {s}\in{Y} \), \( \displaystyle {{y}}^{{-{1}}}{{a}}^{{-{1}}}={T}{\left({a}{y}\right)}={T}{\left({a}\right)}{T}{\left({y}\right)}={{a}}^{{-{1}}}{{y}}^{{-{1}}} \), i.e. \( \displaystyle {y}\in{C}{\left({a}\right)} \) (with \( \displaystyle {C}{\left(\cdot\right)} \) we've indicated the centralizer of \( \displaystyle \cdot \)).

Now remember that \( \displaystyle {C}{\left(\cdot\right)} \) is a subgroup, and in particular we have just shown that \( \displaystyle \forall{a}\in{Y} \), \( \displaystyle {\left|{C}{\left({a}\right)}\right|}\gt\frac{{\left|{G}\right|}}{{2}} \); by Lagrange's theorem, \( \displaystyle {C}{\left({a}\right)}={G} \). This means that every element of \( \displaystyle {Y} \) commutes with all the element of the group: in other words, \( \displaystyle {Y}\subseteq{Z}{\left({G}\right)} \). But \( \displaystyle \frac{{3}}{{4}}{\left|{G}\right|}\lt{\left|{Y}\right|}\le{\left|{Z}{\left({G}\right)}\right|} \) which gives \( \displaystyle {Z}{\left({G}\right)}={G} \) so \( \displaystyle {G} \) is abelian.

In this case, we can easily prove that \( \displaystyle {Y} \) is indeed a subgroup and, since it has got more than \( \displaystyle \frac{{3}}{{4}} \) of the elements of \( \displaystyle {G} \), it must be \( \displaystyle {G} \) itself, i.e. \( \displaystyle {T}{x}={{x}}^{{-{1}}},\forall{x}\in{G} \).

Question. What if \( \displaystyle {T}{x}={{x}}^{{-{1}}} \) for exactly \( \displaystyle \frac{{3}}{{4}} \) of the elements of \( \displaystyle {G} \)?