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Definition. A finite group $G$ is called squared (resp. almost squared) if there exists a subset $A\subseteq G$ such that $G=\{ab:a,b\in A\}$ and $|G|=|A|^2$ (resp. $|G|=|A|^2-1$). Such a set $A$ will be called an (almost) square root of $G$.

According to the answers to this MO-post, no nontrivial finite group is squared.

In contrast, nontrivial almost squared finite groups do exist. The simplest one is the 3-element cyclic group $C_3$. Any 2-element subset of $C_3$ is an almost square root of $C_3$.

A less trivial example is the dihedral group $D_8=\langle a,b\;|\;a^4=b^2=1,\;bab=a^3\rangle$ with almost square root $A=\{a,b,ba\}$.

Three other examples of almost squared groups (found by GAP) are:

$\bullet$ the symmetric group $S_4$;

$\bullet$ the general linear group $GL(2,3)$ of non-degenerate $2\times 2$ matrices over the 3-element field,

$\bullet$ the symmetric group $S_5$.

Problem 1. Find more examples of almost squared finite group. Are there infinitely many almost squared finite groups?

Remark. Using GAP, Voldymyr Gavrylkiv established that among groups of order $<168$ the only almost squared groups are the groups $C_3$, $D_8$, $S_4$, $GL(2,3)$, and $S_5$. Those groups have orders 3, 8, 24, 48, and 120, respectively. It is interesting that no almost squared group of order 80 exists.

Problem 2. What can be said about the structure of almost squared groups?

Remark. Alex Ravsky observed that for an almost square root $A$ of an almost squared group $G$, the center $Z(G)$ of $G$ is almost contained in the set $A^2=\{a^2:a\in A\}$ in the sense that $Z(G)\setminus A^2$ contains at most one element. So, $|Z(G)|\le|A|+1=1+\sqrt{|G|+1}$, which implies that the unique almost squared commutative group is $C_3$.

The only known (at the moment) almost squared noncommutative groups $D_8$, $S_4$ and $GL(2,3)$ have even cardinality.

Problem 3. Is the cardinality of any almost squared noncommutative group even?

Remark. It can be shown that all noncommutative groups of odd order $<675=3^3\times 5^2$ are not almost squared.

Problem 3'. Is there an almost squared group among groups of order $675$?

Problem 4. Let $A$ be an almost square root of an almost squared non-commutative finite group $G$. Are there distinct elements $a,b\in A$ such that $a^2=b^2=g$ for some $g\in G$ such that:
$\bullet$ $g\in Z(G)$?
$\bullet$ $g^2\in Z(G)$?
$\bullet$ $g^2=1$?
$\bullet$ $g=1$?
$\bullet$ $g$ has order $\le 3$?

Remark. For any almost square root in any of three known almost squared noncommutative groups $D_8,S_4,GL(2,3)$, there exist distinct elements $a,b\in A$ such that $a^2=b^2=1$. In these group the center contains at most two elements.

The following problem was suggested by @LSpice in his comment.

Problem 5. Let $G$ be an almost squared noncommutative group. Is $z^2=1$ for any central element of $G$?

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    $\begingroup$ I tried to extend the arguments in the representation ring. The eigenvalues of $\sum_{a \in A} a$ on every nontrivial irreducible representation of $a$ are roots of unity, since the eigenvalues of its square are roots of unity on these representations. Thus $|\operatorname{tr} (\sum_{a \in A} a, \dim V)| \leq \dim V$ for all nontrivial representations, so $\sum_v |\operatorname{tr} (\sum_{a \in A} a, \dim V)| ^2 \leq |A|^2 + |G|-1= 2 (n^2-1)$ $\endgroup$ – Will Sawin Feb 20 at 21:05
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    $\begingroup$ which by orthogonality of characters implies that $A$ is reasonably evenly distributed among the conjugacy classes of $G$ - if I calculated right that the sum over conjugacy classes $C$ of $( | A \cap C| - |A| |C| /|G|)^2 / |C|$ is at most $(n^2-2)/(n^2-1)$, but this doesn't seem strong enough to lead to a classification. $\endgroup$ – Will Sawin Feb 20 at 21:06
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    $\begingroup$ @LSpice Right. Thank you. I corrected to "every". $\endgroup$ – Taras Banakh Feb 21 at 4:46
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    $\begingroup$ Trivial observation: if $A$ is an almost square root of $G$, then so is $z A$ for every $z \in \operatorname Z(G)$; so perhaps it makes sense to ask in Problem 4 if some central translate of $A$ satisfies your condition—unless you are making the additional hypothesis that the centre of an almost squared group has exponent $2$? (One could perhaps also ask about how many almost square roots there are; surely it would be too much to expect that they are all central translates of one another.) $\endgroup$ – LSpice Feb 21 at 5:26
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    $\begingroup$ @LSpice For the almost squared groups $D_8$ and $S_4$ the center indeed has exponent 2. GAP calculations show that $S_4$ has 96 almost roots. The simplest is $\{ a, b, c, ac, bc \}$ where $a,b,c$ are elements of $S_4$ such that $a^2=b^3=c^2=1$, $ba=ab^2$, $(aca)^2=1$, $cb=baca$... GAP shows some strange presentation of $S_4$ (if the group (24,12) in GAP is indeed $S_4$). Maybe I will try to find the almost square root of $S_4$ by hands (using some more convenient presentations). $\endgroup$ – Taras Banakh Feb 21 at 6:46
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Some preliminary comments. I may add more later: Let $G$ be any finite group. If $S$ is a subset of $G$, let $S^{+} = \sum_{s \in S} s$ in the group algebra $\mathbb{C}G$. Suppose that $G$ has an almost square root $A$. If $1 \in A$, then $1 a = a 1 $ for every $a \in A$, so that $|A| = 1 $ and $|G| = 3 $, so suppose from now on that $1 \not \in A$. We have $(A^{+})^{2} = x + G^{+}$ for some $x \in G.$ Notice that $A^{+}$ commutes with $G^{+}$, so that $A^{+}$ commutes with $x$. Hence $xAx^{-1} = A$, so that the set $A$ is invariant under conjugation by $x$. Also we have $(x^{n}A^{+})^{2} = x^{2n+1} + G^{+}$ for every integer $n$, so that $x^{n}A$ is another almost square root for $G$. If the order of $x$ is not a power of $2$, then there is an integer $n$ such that the order of $x^{2n+1}$ is a power of $2$. Hence if $G$ has an almost square root, then we may arrange matters so that the order of $x$ is a power of $2$ (allowing the possibility of order $1$). In particular, we may arrange matters so that $x = 1$ when $G$ has odd order.

Now suppose that $x$ has $2$-power order greater than one. Let $a,b \in A$ with $x = ab.$ If $a,b \not \in C_{G}(x)$ then $x = a^{x}b^{x}$ is the only other factorization of $x$ as a product of two elements of $A$. If $a,b \in C_{G}(x)$ with $a \neq b$ then $x = ab = ba$ are the only expressions of $x$ as a product of two elements of $A$. If $a = b \in C_{G}(x)$ then there is exactly one more expression $c^{2} = x$ with $c \in A$ (since $x$ can't then be a product of two elements of $A$, each outside $C_{G}(x)).$

Now let $H = C_{G}(x)$ and $D = A \cap H.$ Suppose that $ d \in H $ with $d \neq x.$ Then $d = uv$ for some $u,v \in A$ and $u,v$ are unique. Then $d = d^{x} = u^{x}v^{x}$ and $u^{x},v^{x} \in A$. Hence $u = u^{x}$ and $v = v^{x}.$ Thus $u, v \in D$.

Now we have $(D^{+})^{2} = H^{+} \pm x $. If the product is $H^{+} + x,$ then $D$ is an almost square root for $H = C_{G}(x).$

Suppose then $(D^{+})^{2} = H^{+}-x.$ Note that $H$ has even order as $x \in H$. In the regular representation of $H$, the element $H^{+}- x$ has the eigenvalue $|H| - 1$ with multiplicity $1$, and $D^{+}$ has the eigenvector $|D|$ on that eiegenspace. Hence $|H|-1$ is the square of an odd integer. Thus $|H| \equiv 2$ (mod $8$). In that case, $\langle x \rangle$ is a Sylow $2$ subgroup of $H = C_{G}(x) = N_{G}(\langle x \rangle)$, so is also a Sylow $2$-subgroup of $G$. This is a standard group-theoretic argument, but I explain it for non-group theorists : Note that $x$ lies in some Sylow $2$-subgroup $S$ of $G$. Suppose that $|S| > 2.$ If $x \in Z(S)$, then $S \leq C_{G}(x),$ so that $|S|$ divides $|H|$ which is not the case. If $x \not \in Z(S)$, then $\langle x, Z(S) \rangle \leq H$ and $|\langle x, Z(S) \rangle | \geq 2|\langle x \rangle |,$ a contradiction.

However, we can now reach a contradiction. If $G$ is a group of even order such that $A$ is an almost square root for $G$, then $|A|^{2} = |G| + 1$, so that $|G|+1$ is the square of an odd integer. Hence $|G| + 1 \equiv 1$ (mod $8$), so that $|G|$ is divisible by $8$.

But in the present case, $|G|$ is not divisble by $4$. Hence the situation $(D^{+})^{2} = H^{+} - x$ can not occur when $x$ is a non-identity element of $2$-power order of $G$, and $G$ has an almost square root $A$ such that $(A^{+})^{2} = G^{+} + x$, where $H = C_{G}(x).$

In particular, we may deduce that if $G$ is a finite group of even order with an almost square root $A$ such that $(A^{+})^{2} = G^{+} + x$ for some non-identity element $x$ whose order is a power of $2$, then $D = A \cap C_{G}(x)$ is an almost square root for $H = C_{G}(x).$

To recap : if $G$ is a finite group with an almost square root $A$, and we have $(A^{+})^{2} = G^{+} + x$ for some non-identity element $x$, then we can find an integer $n$ such that $x^{n}A$ is an almost square root for $G$ with $(x^{n}A^{+})^{2} = G^{+} + x^{2n+1}$, where the order of $x^{2n+1}$ is a power of $2$ (possibly $1$). This reduces to considering the case that the order of $x$ is a power of $2$ (possibly $1$). If the order of $x$ is a power of $2$ greater than one, the group $C_{G}(x)$ has the almost square root $D = A \cap C_{G}(x).$

Hence a key case to understand is when $x$ is a central element of $G$ whose order is a power of $2$ (possibly $1$).

Continued : Notice also that if $x$ has even order, then $|C_{G}(x)|$ is a group of even order with an almost square root, so that $|C_{G}(x)|$ has order divisible by $8$.

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  • $\begingroup$ If $x=ab$ for some $a,b\in A\setminus C_G(x)$, then $xax^{-1}=x^{-1}ax$ and $xbx^{-1}=x^{-1}bx$ and hence $a,b\in C_G(x^2)$. Maybe this can be used somehow? For example that $A\cap C_G(x^2)$ is an almost square root of the subgroup $C_G(x^2)$? $\endgroup$ – Taras Banakh Feb 21 at 17:28
  • $\begingroup$ If $(D^+)^2=H^+-x$, then $x^2=1$. Indeed, in this case $|H|=|D|^2+1$ and $|H|$ is even since $x\in H$ has 2-power order. Then $|D|=2m+1$ for some $m$ and hence $|H|=|D|^2+1=4(m^2+m)+2$ is not divisible by 4. So, the order of $x$ should be 2. $\endgroup$ – Taras Banakh Feb 21 at 21:45
  • $\begingroup$ I noticed (even on the answers to my preceding "squared" question) that there is a kind of telepathy when different mathematicians on huge distances arrive to the same conclusions in the same time. By the way, it is not the end of the story. If $(D^+)^2=H^+-x$, then $x\in D$ and $H=C_G(x)$ is a 2-element group. Assuming that $x\notin D$, we can find an element $u\in D$ with $u^2=1$. Since $u$ and $x$ commute and $x^2=1$, they generate a 4-element subgroup of $H$, which implies that $|H|$ is divisible by 4, which is not true. Therefore $x\in D$. $\endgroup$ – Taras Banakh Feb 21 at 22:06
  • $\begingroup$ Therefore $x\in D$ and since elements of $D$ do not commute, we conclude that $|D|=1$ and $H=\{1,x\}$. Therefore, if $(D^+)^2=H^+-x$, then $x\in D$, $x^2=1$ and $C_G(x)=\{1,x\}$. $\endgroup$ – Taras Banakh Feb 21 at 22:11
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    $\begingroup$ I can also prove that $G$ is even and $a^2=b^2$ for two distinct $a,b\in A$, then the center of $G$ has even index in $G$. All non-commutative examples (i.e., $D_8$, $S_4$ and $GL(2,3)$ of almost squared groups show that exacty this happens: any almost square root contains two distinct elements $a,b$ with $a^2=b^2=1$. $\endgroup$ – Taras Banakh Feb 23 at 8:51
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The following theorem yields a partial answer to Problem 2.

A subset $C$ of a group $G$ is called unfree if $xy=yx$ or $x^2=y^2$ for any elements $x,y\in C$.

For a group $G$ let $ucov(G)$ be the smallest cardinality of a cover of $G$ by unfree subsets of $G$.

Theorem 1. If $A$ is an almost square root of an almost squared finite group $G$, then $\sqrt{|G|+1}=|A|\le 1+ucov(G).$$

Proof. Let $A$ be an almost square root of $G$ and $\mathcal U$ be a cover of the group $G$ by unfree subsets such that $|\mathcal U|=ucov(G)$. We lose no generality assuming that the cover $\mathcal U$ consists of pairwise distinct sets.

Claim. There exists an element $a\in A$ such that for any distinct elements $x,y\in A\setminus\{a\}$ we have $xy\ne yx$ and $x^2\ne y^2$.

Proof. To derive a contradiction, assume that for any $a\in A$ there exist elements $x,y\in A\setminus\{a\}$ such that $xy=yx$ or $x^2=y^2$. For every $g\in G$ consider the set $P_g=\{(a,b)\in A\times A:ab=g\}$. By the definition of an almost square root, there exists a unique element $g\in G$ such that $|P_g|=2$ and $|P_x|=1$ for all $x\in G\setminus\{g\}$. Take any pair $(a,b)\in P_g$. By our assumption, there exist two distinct elements $x,y\in A\setminus\{a\}$ such that $xy=yx$ or $x^2=y^2$. It follows that the set $P_g$ coincides with $\{(x,y),(y,x)\}$ or $\{(x,x),(y,y)\}$, which implies that $a\in\{x,y\}$. But this contradicts the choice of $x,y$. $\quad\square$

Claim implies that for any $U\in\mathcal U$ we have $|U\cap (A\setminus\{a\})|\le 1$. Consequently, $$|A|=1+|A\setminus\{a\}|\le 1+|\mathcal U|=1+ucov(G).\quad\square$$

Now it remains to find some upper bounds on the number $ucov(G)$.

For a group $G$ let $ccov(G)$ be the smallest cardinality of a cover of $G$ by cyclic subgroups. Since each cyclic group is an unfree set, we get the upper bound $ucov(G)\le ccov(G)$. Also $ccov(G)\le|G|-1$ for any nontrivial group $G$ and $ccov(G)\le|G|-2$ for any group which is not Boolean.

Lemma 1. $ucov(G)\le ccov(G/Z(G))$.

Proof. Let $Z=Z(G)$ be the center of $G$. Let $\kappa=ccov(G/Z)$ and $(C_\alpha)_{\alpha\in\kappa}$ be a cover of $G/Z$ by cyclic subgroups. For every $\alpha\in\kappa$ choose an element $c_\alpha\in G$ such that the coset $c_\alpha Z$ is a generator of the cyclic subgroup $C_\alpha$. Observe that for every $\alpha\in\kappa$ the set $U_\alpha=\bigcup_{n\in\mathbb Z}c_\alpha^nZ$ is unfree and $\mathcal U=\{U_\alpha\}_{\alpha\in\kappa}$ is a cover of $G$ by unfree sets withessing that $$ucov(G)\le|\mathcal U|\le\kappa=|ccov(G/Z)|.\quad$$

Corollary. If a finite noncommutative group $G$ is almost squared, then $|Z(G)|\le \sqrt{|G|+1}-1$.

Proof. Let $A$ be an almost square root of $G$. By Theorem 1 and Lemma 1, we have $$|A|\le 1+ucov(G)\le 1+cc(G/Z(G))\le 1+|G/Z(G)|-1=|G|/|Z(G)|$$ and hence $$|Z(G)|\le \frac{|G|}{|A|}=\frac{|A|^2-1}{|A|}=|A|-\frac1{|A|}$$ and finally $$|Z(G)|\le -1+|A|=-1+\sqrt{|G|+1}.$$

Proposition. If a non-commuattive group $G$ of order 120 is almost squared, then $|Z(G)|\le 6$.

Proof. Let $Z=Z(G)$ be the center of the group $G$ and $A$ be an almost square root of $G$. By the preceding Corollary, $|Z|\le 10$. Assuming that $|Z|=10$, we conclude that the group $G/Z$ has order 12. Looking at the classification of groups of order 12, we can observe that $G/Z$ contains a cyclic subgroup of order $\ge 4$. Then $ccov(G/Z)\le 1+(12-4)=9$ and hence $11=|A|\le 1+ucov(G)\le 1+ccov(G/Z)\le 10$, which is not true. This contradiction shows that $|Z|<10$ and hence $|Z|\le 8$.

Assuming that $|Z|=8$, we conclude that the group $G/Z$ has order $15$ and hence is cyclic, which implies $11=|A|\le 1+ucov(G)\le 1+ccov(G/Z)=2$ and this is a contradiction. So, $|Z|<8$ and hence $|Z|\le 6$ (as $|Z|$ divides $|G|=120$).

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