I was helping a friend prepare for his intro abstract final and he mentioned the professor had once asked the question: name a group and an automorphism that takes $3/4$ of the elements of the group to their own inverses (for instance, the dihedral group $D_4$ of order $8$, with identity automorphism). I tried to figure out how to approach this question in general but can't see how.

1) Can we construct all such groups?

(it is asserted in comment to the answer below that these are precisely those finite groups whose center has index 4)

2) Given a rational number $a/b\in [0,1]$ does there exist a finite group $G$ and an automorphism $f$ such that $f$ maps exactly $a/b$ elements of $G$ to their own inverses?

($a/b=1$ is achieved precisely for the inversion map on an abelian group; otherwise $a/b\le 3/4$ according to the answer below)

3) Also, can these questions make sense in infinite groups?

  • $\begingroup$ Maybe edit the title of your question: a/b is a rational number $\le 1$. And if we allow infinite groups, how do you define a fraction of the elements? Unless there is a natural density function, you might be in trouble, for example: $(\mathbb{Z}/2)^\mathbb{N} \times \mathbb{Z}^\mathbb{N}$ $\endgroup$ Jan 3 '12 at 23:14

This may be a well-known chestnut? (well-known to those that know it well, that is)

The fraction can never be between 3/4 and 1. To prove this, suppose $\phi\colon G\to G$ is an automorphism of $G$ that sends more than 3/4 of the elements of $G$ to their inverses. Let $S=\lbrace g\in G\colon \phi(g)=g^{-1}\rbrace$.

Notice that if $g$, $h$ and $gh$ all lie in $S$ then on the one hand, $\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$. On the other hand, $\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$, so that $g^{-1}$ and $h^{-1}$ commute. It follows that $g$ and $h$ commute.

Fix $g\in S$ and consider $A=\lbrace h\colon h\in S\text{ and } gh\in S\rbrace$. There are less than $|G|/4$ $h$'s for which the first condition fails and less than $|G|/4$ $h$'s for which the second condition fails, so that $|A|>|G|/2$. By the above, it follows that the centralizer of $g$ (i.e. the set of $h$'s that commute with $g$) is a superset of $A$. Since the centralizer is a subgroup, by Lagrange's theorem the centralizer of $g$ must be all of $G$. That is $g$ lies in the center of the group. Now we have that more than $1/2$ of the group lies in the center (which is again a subgroup), so that $G$ is Abelian.

Now since $S$ is more than half of the group, the subgroup generated by $S$ must be all of $G$, so that every element of $G$ is a product of elements of $S$. Now $g\in G$, write $g=s_1\ldots s_n$. Then $\phi(g)=\phi(s_1)\ldots\phi(s_n)=s_1^{-1}\ldots s_n^{-1}=s_n^{-1}\ldots s_1^{-1}=g^{-1}$, so that $\phi(g)=g^{-1}$ for all $g\in G$.

  • 7
    $\begingroup$ I believe the full set of fractions is known, see MacHale and Liebeck, "Groups With Automorphisms Inverting Most Elements". $\endgroup$
    – Steve D
    Jan 4 '12 at 0:49
  • 1
    $\begingroup$ This answer shows that the ratio $|S|/|G|$ can never lie strictly between $3/4$ and $1$, but the ratio $3/4$ itself is actually possible. Namely, the dihedral group of order $8$ has an automorphism that sends precisely $6$ elements to their respective inverses. $\endgroup$ Jan 4 '12 at 0:53
  • 5
    $\begingroup$ Yes, in fact the paper I mentioned above shows the only possibilities between 1/2 and 1 are fractions of the form $(n+1)/2n$, for integers $n$. Every such fraction is realized, and if a group realizes one fraction, it cannot realize another. $\endgroup$
    – Steve D
    Jan 4 '12 at 0:55
  • 2
    $\begingroup$ Also groups realizing $3/4$ are precisely those with center of index $4$. $\endgroup$
    – Steve D
    Jan 4 '12 at 0:56
  • $\begingroup$ Yes, this is a well-known chestnut: mathoverflow.net/questions/48 . $\endgroup$ Dec 15 '12 at 6:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.