# How many square roots can a non-identity element in a group have?

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.

• 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 $4$. Dec 13, 2021 at 13:29
• 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. Dec 13, 2021 at 13:46
• 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. Dec 15, 2021 at 16:32

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 $$G$$.

• 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? Dec 16, 2021 at 0:55
• @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 $C_G(y)=G$. Dec 16, 2021 at 6:38
• Thank you! I essentially copied this argument verbatim, see Proposition 3.12 in arxiv.org/pdf/2204.09666.pdf Apr 21, 2022 at 10:37
• @alpmu Thanks for the reference! Apr 21, 2022 at 11:33

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 $$r_2(g)=6$$.

• (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 ... Dec 13, 2021 at 16:05
• ... 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. Dec 13, 2021 at 16:30
• True! I saw a way to improve the argument a bit but not enough to reach $\frac{3}{4}$. Dec 14, 2021 at 23:27
• @EmilJeřábek See my streamlined and simplified version, based on your posts and the original post above. Dec 15, 2021 at 6:58

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.

• 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. Dec 13, 2021 at 19:48
• 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. Dec 13, 2021 at 19:49
• @WillSawin Yes that's a good way to finish the argument! Dec 13, 2021 at 21:54

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 $$r_2(g) = |G|/2$$.

• 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. Dec 13, 2021 at 20:14
• 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 $\varepsilon^2$). Dec 13, 2021 at 21:06
• The official name for your last example is "generalized dicyclic group". Dec 14, 2021 at 6:42