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In the thread Involutions and Abelian Groups, I supplied a solution to the following interesting problem (with the help of hints provided by the OP).

Let $ G $ be a finite group and $ I(G) $ the set of involutions of $ G $ (an involution of $ G $ is defined as an element $ x $ of $ G $ with order $ 2 $, i.e., $ x^{2} = e $). Prove that if $ \dfrac{|I(G)|}{|G|} \geq \dfrac{3}{4} $, then $ G $ must be an abelian group.

This problem naturally led me to consider the following intriguing question.

Can we find non-abelian finite groups $ G $ for which the ratio $ \dfrac{|I(G)|}{|G|} $ is less than but arbitrarily close to $ \dfrac{3}{4} $?

I am pretty sure that this problem has been studied extensively (and exclusively) by group theorists, but I have not been very successful in finding sources that discuss it. I do not even know if this question has an affirmative or negative answer. I therefore welcome all helpful suggestions and insights. Thank you!

Community
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Haskell Curry
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2 Answers2

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Consider the dihedral group $D_8$ of order $8$. Notice that it has $2$ elements of order $4$, $5$ elements of order $2$ and an identity element. Now consider the group $(\mathbb Z_2)^n \times D_8$. This is still non-abelian and we have that there are $2^n \cdot 5+2^n-1$ elements of order $2$ in particular we see that

$$\lim_{n\rightarrow \infty} \frac{6\cdot 2^n-1}{8\cdot 2^n}=3/4.$$

JSchlather
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  • Thank you, Jacob, for an elegant solution. – Haskell Curry Jan 16 '13 at 08:07
  • @HaskellCurry I originally tried $(\mathbb Z_2)^n \times S_3$ but there the ratio goes to $2/3$. I'm curious now if it's possible to find a similar family of groups that aren't $2$-groups. – JSchlather Jan 16 '13 at 08:16
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    I think that $2/3$ is best possible outside of $2$-groups. – Alexander Gruber Jan 16 '13 at 08:26
  • @Alexander: You mentioned in the linked thread that you are interested in such problems. Maybe you can help me refresh my memory. I think that there is the following somewhat similar result. Let $ G $ be a finite group. If there exists a group automorphism $ \sigma: G \to G $ such that $ \sigma(x) = x^{-1} $ for at least three-fourths of group elements $ x $, then $ G $ must be an abelian group. Did I state the result correctly? – Haskell Curry Jan 16 '13 at 08:59
  • @HaskellCurry If $\sigma(x)=x^{-1}$ for $3/4$ of the group then $\sigma^2(x)=x$ for $3/4$ of the group, so the subgroup fixed by $\sigma^2$ is $G$ and thus $\sigma(x)=x^{-1}$ for the whole group. Then it follows because a group $G$ is abelian if and only if the map $\sigma(x)=x^{-1}$ is a homomorphism. – JSchlather Jan 16 '13 at 09:04
  • Yes, it must be an elementary abelian $2$-group. – Alexander Gruber Jan 16 '13 at 09:16
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Here is a proof of my claim in the comments.

If $G$ is a nonabelian group of even composite order with $|I(G)|/|G|>1/2$, then $G$ is generalized dihedral by a theorem of Wall. By a corollary of Frobenius-Schur (which appears as Corollary 4.6 in Isaacs Character Theory of Finite Groups), the number of involutions in $G$ is equal to $$-1+\sum_{\chi\in \text{Irr}(G)} \chi(1).$$ Our strategy will be to count the irreducible characters of each dimension using what we know about the structure of generalized dihedral groups, and thus determine the number of involutions in $G$.

Suppose that $G=A\rtimes \langle \tau \rangle$, where $A$ is abelian. Then the quotient of $A$ by $N:=\{a^2|a\in A\}$ is an elementary abelian $2$-group. Let $k=\text{rank}(A/N)$. Each of the $2^k$ such characters from $A/N$ can send $\tau$ to either $\pm 1$, so Then $G$ has $2^{k+1}$ linear characters, as . Next we can induce the characters of $A$ for which $A_2\not \leqslant \text{ker}(\chi)$ to obtain $(n-2^k)/2$ distinct characters of dimension $2$.

Letting $|A|=2^jm$ for some odd $m>2$, we have that $|G|=2^{j+1}m$, from which it follows that $$\frac{1+|I(G)|}{|G|}=\frac{1}{2^{j+1}m}\left(2^{k+1}+2\cdot \frac{(2^jm-2^k)}{2}\right)=\frac{1}{2}+\frac{1}{2^{1+j-k}m}$$ which is bounded above by $$\frac{1}{2}+\frac{1}{2m}$$ with equality when $j=k$, i.e. when the Sylow $2$-group of $A$ is elementary abelian. This upper bound is a decreasing function of $m$, so the above calculation verifies that $|I(G)|/|G|$ approaches a maximum when we choose $m=3$, whereupon $$\frac{1}{2}+\frac{1}{2m}=\frac{2}{3}.$$ So Jacob's example $\mathbb{Z}_2^k\times S_3\cong D(\mathbb{Z}_3\oplus \mathbb{Z}_2^k)$ is in fact the maximal construction.

Alexander Gruber
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  • Why is $A^\ast$ elementary abelian $2$-group? – JSchlather Jan 16 '13 at 15:19
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    That line should have been set of squares, not involutions. – Alexander Gruber Jan 16 '13 at 15:27
  • @Alexander: Thanks! I like your style. You actually took the time to justify your claim. +1! – Haskell Curry Jan 16 '13 at 18:15
  • @AlexanderGruber: Hi Alexander. I’m very impressed by your solution, but I have trouble understanding certain points. I checked Corollary 4.6 of Isaacs’ book, and the formula for the number of involutions of $ G $ given there is $$ -1 + \sum_{\chi \in \text{Irr}(G)} {v_{2}}(\chi) ~ \chi(1), $$ where $ {v_{2}}(\chi) \in { -1,0,1 } $. I was thus wondering what allows you to take $ {v_{2}}(\chi) = 1 $. I also don’t understand why characters on $ A / N $ give rise to linear characters on $ G $, of which there are a total of $ 2^{k + 1} $, where $ k = \text{Rank}(A / N) $. – Berrick Caleb Fillmore Oct 06 '14 at 22:32
  • By ‘$ A_{2} $’ above, do you mean the alternating group of order $ 2 $? I’d like to know why the ‘$ A_{2} \nleq \ker(\chi) $’ condition is so crucial to inducing characters, and I’m not sure how you obtained $ \dfrac{2^{j} m - 2^{k}}{2} $ as the number of characters of $ G $ of degree $ 2 $. Finally, is there a reason why we aren’t counting characters of degree $ \geq 3 $? I apologize if I sound a little demanding. I just want to understand a beautiful piece of mathematics, that’s all. – Berrick Caleb Fillmore Oct 06 '14 at 22:42