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Let $G$ be a finite group. Let $r_2\colon G \to \mathbb{N}$ be the square-root counting function, assigning to each $g\in G$ the number of $x\in G$ with $x^2=g$. Perhaps surprisingly, $r_2$ does not necessarily attain its maximum at the identity for general groups, see Square roots of elements in a finite group and representation theory.

I'm interested in whether $r_2(g)$ can attain a value above $0.999|G|$ for some non-identity element $g\in G$.

Update: Thanks to everybody who participated in the discussion. The lemma proved here influenced greatly the statement of Theorem 4.2 in https://arxiv.org/pdf/2204.09666.pdf . Proposition 3.12 in this same paper is essentially the answer posted by GH from MO.

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    $\begingroup$ I think not. If $g$ is such an element, then $g$ is equal to all of its conjugates, so $g \in Z(G)$, and $g^2=1$, or else $g^{-1}$ would be another such element. Then a high proportion of the elements of $G/\langle g \rangle$ have order $2$ which for a high enoguh proportion) forces $G/\langle g \rangle$ to be elementary abelian, in which case $G$ is a central product of an abelian group with an extraspecial group. For the extraspecial groups, the proportion of elements of order $4$ can be bigger than $1/2$ but not by much - the largest is for $Q_8$, with $3/4$ of the elements of order $4$. $\endgroup$
    – Derek Holt
    Dec 13, 2021 at 13:29
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    $\begingroup$ You could take one of the central factors to be $Z(G)$, and the other to be the inverse image in $G$ of a complement of $Z(G)/\langle g \rangle$ in $G/\langle g \rangle$. I will expand my comment to an answer later. $\endgroup$
    – Derek Holt
    Dec 13, 2021 at 13:46
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    $\begingroup$ The 3/4 bound is also attained (apart from the $g \neq 1$ requirement, which is not very natural) by the other nonabelian group of order 8. $\endgroup$ Dec 15, 2021 at 16:32

4 Answers 4

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Here is a streamlined and simplified version of the posts by Saúl Rodríguez Martín and Emil Jeřábek.

Theorem. Assume that $G$ is a finite group, and $r_2(g)>(3/4)|G|$ holds for some $g\in G$. Then $G$ is an elementary abelian $2$-group, and $g$ is the identity element.

Proof. Fix any element $y\in G$, and consider the sets $$S=\{x\in G: x^2=g\},\qquad T=\{x\in S:xy\in S\}.$$ By the union bound, $$|G\setminus T|\,\leq\, 2|G\setminus S|<|G|/2,$$ hence $|T|>|G|/2$. For any $x\in T$, we have $(xy)^2=x^2$, which implies that $$xyx^{-1}=(xy)x^{-1}=(xy)^{-1}x=y^{-1}.$$ So $\{x\in G:xyx^{-1}=y^{-1}\}$ contains more than half of the elements of $G$, whence it contains all elements of $G$. In particular, $y=y^{-1}$, which shows that $G$ is an elementary abelian $2$-group. Moreover, $g$ is the identity element, since the identity element is the only square in $G$.

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    $\begingroup$ What is the justification for "contains more than half of the elements of G, whence it contains all elements of G"? Is that set closed under inverses or something? $\endgroup$ Dec 16, 2021 at 0:55
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    $\begingroup$ @MarioCarneiro: The set is a coset of $C_G(y)$, the centralizer of $y$. Now $C_G(y)$ is a subgroup of $G$, hence $|C_G(y)|>|G|/2$ forces $C_G(y)=G$. $\endgroup$
    – GH from MO
    Dec 16, 2021 at 6:38
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    $\begingroup$ Thank you! I essentially copied this argument verbatim, see Proposition 3.12 in arxiv.org/pdf/2204.09666.pdf $\endgroup$
    – alpmu
    Apr 21 at 10:37
  • $\begingroup$ @alpmu Thanks for the reference! $\endgroup$
    – GH from MO
    Apr 21 at 11:33
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Here is an elementary way to prove that there can´t be a finite group $G$ and non-identity $g\in G$ with $r_2(g)>\frac{5}{6}|G|$. Suppose that happens and call $S=\{x\in G; x^2=g\}$. Then of course there must be $x\in G$ with $x,x^{-1}\in S$, so $g^2=1$.

Now for every $x\in G$, $S\cap x^{-1}S$ has more than $\frac{2|G|}{3}$ elements. If for each $y\in G$ you consider the set $A_y=\{x\in G;y\in S\cap x^{-1}S\}$, then $\sum_{y\in G}|A_y|=\sum_{x\in G}|S\cap x^{-1}S|>|G|\frac{2|G|}{3}$, so there is some $y$ with $|A_y|>\frac{2|G|}{3}$. So, $|S\cap A_y|>\frac{|G|}{2}$. Pick $x\in S\cap A_y$.

Then we have $y^2=xyxy=x^2=g$.

From these equalities we deduce $xyx=y$, and as $x^2y^2=g^2=1$, we have $xy=x^{-1}y^{-1}$. So, $y=xyx=x^{-1}y^{-1}x$. This means that for all possible choices of $x$, which is more than $\frac{|G|}{2}$, $x^{-1}y^{-1}x= y$. So, $x^{-1}y^{-1}x=y$ for all $x$, which is impossible since $y\neq y^{-1}$.

Edit: As Emil Jeřábek points out in the comments, this argument can be refined to prove that $r_2(g)>\frac{3}{4}|G|$ can´t be achieved. The bound $r_2(g)=\frac{3}{4}|G|$ is reached in the example Derek Holt mentions in his answer: the group $Q_8$ and its element $g$ with $r_2(g)=6$.

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    $\begingroup$ (My original comment was incorrect.) If you assume only $r_2(g)>\frac34|G|$, your argument gives $y$ with $|S\cap A_y|>\frac14|G|$, thus $|\{x:x^{-1}yx=y^{-1}\}|>\frac14|G|$. This is a coset of $C(y)$, hence $[G:C(y)]\le3$; in fact, $y^{-1}\ne y$ and $x^{-2}yx^2=y$ implies $[G:C(y)]=2$, and $S\cap A_y$ is the nontrivial coset of $C(y)$. Now I’m not sure how to finish the argument ... $\endgroup$ Dec 13, 2021 at 16:05
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    $\begingroup$ ... Oh, I see: if $w\in C(y)$, then $w\in S$ iff $(y^{-1}w)^2=1$, thus $|C(y)\cap S|=|\{w\in C(y):w^2=1\}|$, which are disjoint, thus $|C(y)\cap S|\le\frac12|C(y)|$ and $|S|\le\frac34|G|$, a contradiction. $\endgroup$ Dec 13, 2021 at 16:30
  • $\begingroup$ True! I saw a way to improve the argument a bit but not enough to reach $\frac{3}{4}$. $\endgroup$
    – Saúl RM
    Dec 14, 2021 at 23:27
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    $\begingroup$ @EmilJeřábek See my streamlined and simplified version, based on your posts and the original post above. $\endgroup$
    – GH from MO
    Dec 15, 2021 at 6:58
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I am just turning my comment into an answer. The answer to the question is no. I think the highest possible value of $r_2(g)$ with $g \ne 1$ is $r_2(g) = 3/4$ for the central element $g$ of $Q_8$, but I haven't proved that formally.

In general, suppose that $g \ne 1$ and $r_2(g) > |G|/2$. Then since conjugate elements of $G$ have the same value of $r_2$, we must have $g \in Z(G)$. We also have $g^2=1$, since otherwise we would have $r_2(g) = r_2(g^{-1})$.

So the proportion of elements of $G/\langle g \rangle$ with $g^2 = 1$ is at least $r_2(G)$. Now, it is proved here that if at least $3/4$ of the elements of finite group $H$ satisfy $g^2=1$ then $H$ is an elementary abelian $2$-group, so this applies to $G/\langle g \rangle$.

Now $G$ is a central product of $Z := Z(G)$ with the inverse image $E$ in $G$ of a complement $E/\langle g \rangle$ of $Z/\langle g \rangle$ in $G/\langle g \rangle$. Then clearly $Z(E) = [E,E] = \langle g \rangle$, so $E$ is an extraspecial $2$-group. Note also that $Z$ is either elementary abelian or it is the direct product of an elementary abelian group with $C_4$.

Now the elements of order $2$ and $4$ in extraspecial groups correspond to the number of elements with $Q(x)=0$ or $1$ in a quadratic form over ${\mathbb F}_2$. We get the highest number of order $4$ in extraspecial groups of minus-type, where the proportion, for a group of order $2^{1+2k}$, is $1/2 + 1/2^{k+3}$, which gives a maximum of $3/4$ when $k=1$.

It seems clear that taking the central product with $Z$ will not change this proportion significantly - in fact it seems to reduce it for the minus-type group.

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  • $\begingroup$ I think the statement about $Q$ is true without the extraspecial assumption. $G$ is an extension of an elementary abelian 2-group $V$ by $\mathbb Z/2$, and these are classified by $H^2(V, \mathbb Z/2)$, which is the space of quadratic forms on $V$. So we're just interested in, for all quadratic forms over vector spaces in characteristic $2$, the one that takes the value $1$ the most often. $\endgroup$
    – Will Sawin
    Dec 13, 2021 at 19:48
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    $\begingroup$ If $x$ in the vector space satisfies $B(x,y)=0$ for all $y$, with $B$ the associated bilinear form, and $Q(x)=1$, then $Q(x+y) = Q(x) + Q(y)$ so exactly one of $Q(x+y)$ and $Q(y)=1$, so the proportion is $1/2$. Thus if $B(x,y)=0$ for all $y$ we may assume $Q(y)=0$, allowing us to reduce inductively to the nondegenerate case, where your calculation applies, finishing the argument and showing $3/4$ is optimal. $\endgroup$
    – Will Sawin
    Dec 13, 2021 at 19:49
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    $\begingroup$ @WillSawin Yes that's a good way to finish the argument! $\endgroup$
    – Derek Holt
    Dec 13, 2021 at 21:54
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In principle one can get an exact answer to this question by a lot of case checking and a quantitative version of the following result.

Theorem. Let $G$ be a finite group such that $r_2(g) \geq \varepsilon |G|$ for some $g \in G$ and $\varepsilon > 0$. Then $G$ contains a subgroup $H$ of index $O_\varepsilon(1)$ which is $2$-step nilpotent with $|[H,H]| \ll_\varepsilon 1$.

Roughly speaking, this says that a group $G$ with a large value of $r_2$ must be ``bounded-by-abelian-by-bounded'', which should be enough structure to then do a case analysis, at least in principle.

Proof. Since all conjugates of $g$ have a value of $r_2$ greater than or equal to $\varepsilon |G|$, there are at most $1/\varepsilon$ conjugates of $g$. Thus the centraliser $C_G(g)$ of $g$ has index $O_\varepsilon(1)$ (cf., Derek Holt's comment). By the pigeonhole principle, there is a coset $x C_G(g)$ of $C_G(g)$ which contains $\gg_\varepsilon |G|$ solutions $xt, t \in C_G(g)$ to the equation $(xt)^2 = g$; by shifting $x$ without loss of generality we may also assume that $x^2=g$. From $x^2 = g$ we have $x = x^{-1} g$ and hence $(xt)^2 = xtx^{-1} gt = xtx^{-1} tg$ since $t \in C_G(g)$. Thus the equation $(xt)^2=g$ is equivalent to $$xtx^{-1} = t^{-1}.\quad (1)$$ So (1) holds for $\gg_\varepsilon |G|$ choices of $t \in G$. In particular, for $\gg_\varepsilon |G|^2$ pairs $(t,s) \in G^2$, we have $$ xtx^{-1} = t^{-1}; \quad xts x^{-1} = s^{-1} t^{-1}$$ which implies $$ xsx^{-1} = t s^{-1} t^{-1}.$$ By Cauchy-Schwarz, this implies that there are $\gg_\varepsilon |G|^3$ triples $(s,t_1, t_2)$ such that $$ t_1 s^{-1} t_1^{-1} = t_2 s^{-1} t_2^{-1}$$ and so there are $\gg_\varepsilon |G|^2$ solutions to the equation $xyx^{-1} = y$. In other words, the commuting probability of $|G|$ is $\gg_\varepsilon 1$. The claim then follows from a theorem of Neumann (see Theorem 2.4 of this paper of Eberhard). $\Box$

I suspect that in the case $\varepsilon > 1/2$ one can work a little harder and show that one can assume without loss of generality that the abelianisation $H/[H,H]$ is $2$-torsion, so that $G$ is ``virtually'' a $2$-torsion group in some weak sense. On the other hand, once $\varepsilon$ reaches $1/2$ one can have non-$2$-torsion behavior. For instance, let $H$ be an arbitrary abelian group, let $g$ be an arbitrary element of $H$ [EDIT: as pointed out by Will below, one needs to assume that $g$ is of order $2$], and consider the group $\langle H, e \rangle$ where $e$ is subject to the relations $e^2 = g$, $ehe^{-1} = h^{-1}$ for all $h \in H$ (this is a sort of twisted dihedral group, I don't know the official name for it). Then $G = \langle H,e \rangle$ is a group containing $H$ as an index $2$ subgroup with $r_2(g) = |G|/2$.

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    $\begingroup$ Your final example works (i.e. contains $H$ as a subgroup) only if $g$ has order $2$, since taking $h=e^2$ we get $e e^2 e^{-1} = e^{-2}$ so $e^4 =1 $ so $g^2=1$. This still doesn't force $H$ to be $2$-torsion, of course. $\endgroup$
    – Will Sawin
    Dec 13, 2021 at 20:14
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    $\begingroup$ This theorem also appears in the paper "Groups satisfying identities with high probability" by Avinoam Mann mathscinet.ams.org/mathscinet-getitem?mr=3899225 As your proof, Mann's proof goes by proving that the commuting probability is large, but this is achieved using characters (a formula by Frobenius-Schur expresses the number of square roots of $g$ in terms of a sum over characters, and a simple application of Cauchy-Schwarz yields that the commuting probability is at least $\varepsilon^2$). $\endgroup$ Dec 13, 2021 at 21:06
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    $\begingroup$ The official name for your last example is "generalized dicyclic group". $\endgroup$ Dec 14, 2021 at 6:42

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