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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).$$

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

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

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

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 $|G|=120$).

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

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

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

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

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 $|G|=120$).

Proposition 168. If a non-commutative group $G$ of order 168 is almost squared, then $|Z(G)|\le 7$.

Proof. Let $Z=Z(G)$ be the center of the group $G$ and $A$ be an almost square root of $G$. By Corollary, $|Z|\le 12$. Assuming that $|Z|=12$, we conclude that the group $G/Z$ has order 14 and hence is cyclic. By Theorem 1, $$13=|A|\le 1+ucov(G)\le 1+ccov(G/Z)=2,$$which is a contradiction showing that $|Z|\ne 12$. Then $|Z|<12$ and hence $|Z|\le 8$ (11,10 and 9 do not divide $168=|G|$). If $|Z|=8$, then the group $G/Z$ has cardinality $168/8=21$ and hence is cyclic. Then $$13=|A|\le 1+ucov(G)\le 1+ccov(G/Z)=2,$$which is a contradiction showing that $|Z|\ne 8$. Therefore, $|Z|\le 7$.

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 $x,y\in A\setminus A\setminus\{a\}$ 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 with 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 $|G|=120$).

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 $|G|=120$).

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 $x,y\in A\setminus A\setminus\{a\}$ 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 with 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. For any almost squared non-commutative group $G$, the center $Z(G)$ has cardinality at most $\sqrt{|G|+1}-1$.

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 $$|Z(G)|\le -1+|A|=-1+\sqrt{|G|+1}.\quad\square$$

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 $x,y\in A\setminus A\setminus\{a\}$ 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 with 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 $|G|=120$).