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kabenyuk
kabenyuk
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If $Z(G)\neq\{e\}$ and $z\in Z(G)$, $z\neq e$, then there are two possible cases, either $z=ab$, $a\neq b$ and $a,b\in A$ or $z=a^2$ where $a\in A$. In the first case we have $z=ab=ba$ and $z^2=a^2b^2=b^2a^2$; but this contradicts the uniqueness of the singular element (if or $a^2=b^2$$z=ab$, then $z^2=a^4$$a\neq b$ and see the second case)$a,b\in A$. In

In the secondfirst case we we have $z=a^2$. Since the order of $G$ is odd, it follows that    $a\in Z(G)$ and $ab=ba$ for each $b\in A$. Hence $|A|=2$ and $|G|=3$.

In the second case we have $z=ab=ba$ and $z$ is singular element. Since there is exactly one singular element, the equality $z^2=a'b'$ with $a',b' \in A$, $a'\neq b'$, is impossible. Then $z^2=a'^2$, $a'\in A$, and we are back to the first case. This is a contradiction and the lemma is proved.

If $Z(G)\neq\{e\}$ and $z\in Z(G)$, $z\neq e$, then there are two possible cases, either $z=ab$, $a\neq b$ and $a,b\in A$ or $z=a^2$ where $a\in A$. In the first case we have $z=ab=ba$ and $z^2=a^2b^2=b^2a^2$; but this contradicts the uniqueness of the singular element (if $a^2=b^2$, then $z^2=a^4$ and see the second case). In the second case we have $z=a^2$. Since the order of $G$ is odd, it follows that  $a\in Z(G)$ and $ab=ba$ for each $b\in A$. Hence $|A|=2$ and $|G|=3$. This is a contradiction and the lemma is proved.

If $Z(G)\neq\{e\}$ and $z\in Z(G)$, $z\neq e$, then there are two possible cases, either $z=a^2$ where $a\in A$ or $z=ab$, $a\neq b$ and $a,b\in A$.

In the first case we have $z=a^2$. Since the order of $G$ is odd, it follows that  $a\in Z(G)$ and $ab=ba$ for each $b\in A$. Hence $|A|=2$ and $|G|=3$.

In the second case we have $z=ab=ba$ and $z$ is singular element. Since there is exactly one singular element, the equality $z^2=a'b'$ with $a',b' \in A$, $a'\neq b'$, is impossible. Then $z^2=a'^2$, $a'\in A$, and we are back to the first case. This is a contradiction and the lemma is proved.

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kabenyuk
kabenyuk
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Simple group of order 360

Another good example of an almost square simple group is the alternating group $A_6$ of order $360=19^2-1$. In order to check that the set $A$ below is an almost square root of $A_6$, you can use the following GAP commands:

g:=AlternatingGroup(6);;
A:=[(1,2)(3,4),(1,3)(5,6),(1,2,3,4)(5,6),(1,2,3),(1,3,4),
(1,2)(3,4,5,6),(1,2)(3,4,6,5),(1,2,3,4,5),(1,2,3,4,6),
(1,2,5)(3,4,6),(1,2,5,3,4),(1,2,5,6)(3,4),(1,2,5,6,3),
(1,2,6)(3,4,5),(1,2,6,3,4),(1,2,6,5)(3,4),(1,2,6,5,3),
(1,3,4,5,6),(1,3,4,6,5)];;
Size(A);
prod:=[];;
for a in A do for b in A do Add(prod,a*b);od;od;    
Size(AsSet(prod));

Note that the subset $$ V=\{(1,2)(3,4),(1,3)(5,6),(1,2,3,4)(5,6),(1,2,3),(1,3,4)\}\subset A $$ is an almost square root of a group isomorphic to the symmetric group $S_4$. The set $V$ is an initial set from which our algorithm starts. We next consider the following three 5-element subsets of $A_6$: \begin{eqnarray*} V_1&=&\{(1,2)(3,4),(1,3)(5,6),(1,2,3,4)(5,6),(1,2,3),(1,3,4)\},\\ V_2&=&\{(1,2)(3,4),(1,3)(5,6),(1,2,3,4)(5,6),(1,3,2),(2,3,4)\},\\ V_3&=&\{(1,2)(3,4),(1,3)(5,6),(1,2,3,4)(5,6),(1,4,2),(2,4,3)\}. \end{eqnarray*} Each of these subsets has the following three properties: (1) $V_i$ is an almost square root of a group isomorphic to $S_4$; (2) $V_i$ has a subset with three elements which is an almost square root of a group isomorphic to the dihedral group $D_4$ of order 8; (3) $V_i$ contains two elements of order 2.

It turns out that any 5-subset of $A_6$ with these three properties is conjugate to one of $V_i$ or $V_i^{-1}$ under the group Aut$(A_6)$. Each set $V_i$ is complemented to an almost square root of $A_6$.

Groups of odd order

It seems to me that the answer to Problem 3' is negative.

Proposition. Any group of order $675$ is not an almost squared group.

Proof. This is a straightforward consequence of Lemmas 1 and 2.

Lemma 1. Let $G$ be a group of odd order and $|G|>3$. If $G$ has an almost square root, then $Z(G)=\{e\}$.

Proof. Let $A$ be an almost square root of $G$. Then there exists exactly one element $x\in G$ which can be represented as the product $x=ab$, $a,\,b\in A$ in exactly two ways. We will call this element a singular element of group $G$ with respect to the almost square root $A$. All the other elements of group $G$ are represented uniquely as the product of two elements of $A$.

If $Z(G)\neq\{e\}$ and $z\in Z(G)$, $z\neq e$, then there are two possible cases, either $z=ab$, $a\neq b$ and $a,b\in A$ or $z=a^2$ where $a\in A$. In the first case we have $z=ab=ba$ and $z^2=a^2b^2=b^2a^2$; but this contradicts the uniqueness of the singular element (if $a^2=b^2$, then $z^2=a^4$ and see the second case). In the second case we have $z=a^2$. Since the order of $G$ is odd, it follows that $a\in Z(G)$ and $ab=ba$ for each $b\in A$. Hence $|A|=2$ and $|G|=3$. This is a contradiction and the lemma is proved.

Lemma 2. Let $G$ be a group of order $675=3^3\cdot5^2$. Then $Z(G)\neq\{e\}$.

Proof. Using the small groups library of GAP we can prove of Lemma 2 by the following GAP commands:

a:=AllSmallGroups(675);;
Filtered(a,x->Size(Center(x))=1);

However, the same can easily be proved without GAP.