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Let $G$ be a group and $$S:=\{g^2 \mid g\in G\}$$ the subset of all squares of $G$. Is $S$ then a subgroup? I would say no since I don't see why $g^2h^2$ should be a square in the non-abelian case. But I didn't find any counter-example. Even if we take $G=S_4$ (a very non-abelian group) I computed $S=A_4$.

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  • $\begingroup$ Of course, the subset of squares of $G$ is not a group. There is only a group generated by the squares. $\endgroup$
    – GAVD
    Jul 24, 2015 at 8:11
  • $\begingroup$ That's what I thought as I pointed out. But give a counter-example then. $\endgroup$ Jul 24, 2015 at 8:18
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    $\begingroup$ It's interesting that for the first few examples for finite groups that one tries, the squares do in fact form a subgroup; the smallest counterexample I know is $A_4$. Here's a good brain-teaser: For which symmetric groups is the set of squares a subgroup? $\endgroup$ Jul 24, 2015 at 8:34
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    $\begingroup$ @JohnBrevik Look here: math.stackexchange.com/questions/1372267/… $\endgroup$ Jul 24, 2015 at 8:35
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    $\begingroup$ @John Brevik: Since abelian groups and dihedral groups do have this property, and so does the quaternion group, the smallest counterexamples indeed have order $12$. In addition to $A_4$, there is also the dicyclic group of order $12$. $\endgroup$
    – verret
    Jul 24, 2015 at 17:34

2 Answers 2

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OK. Let us give some fact:

1) If $G$ is an Abelian group and $H = \{g^2| g \in G\}$, then $H$ is a subgroup. (easy exercise)

2) If $G$ is a dihedral group $D_n$, then the squares forms a subgroup (group of rotations).

So, if you want to take an counter-example, please take a non-abelian group. For example, $G = SL(2,\mathbb{Z}_3)$, the group of $2\times 2$ matrices with entries in $\mathbb{Z}_3$ and determinant 1 (mod 3). Now, you know that there are 24 such matrices. They give 10 squares. Finally, check that

$$\begin{pmatrix}1 &1 \\0 &1\end{pmatrix}^2 \begin{pmatrix}1 &0 \\1 &1\end{pmatrix}^2 = \begin{pmatrix}2 &2 \\2 &1\end{pmatrix},$$ which is not square.

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    $\begingroup$ If you know that there are exactly 10 squares and the group is of order 24 then it can't be a subgroup because $10 \nmid 24$ so you don't need to provide a product of two squares which is no square. $\endgroup$ Jul 24, 2015 at 8:49
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    $\begingroup$ In a dihedral group, the squares do form a subgroup of the group of rotations, but it might not be the full group of rotations. $\endgroup$
    – verret
    Jul 24, 2015 at 17:32
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Take any nonabelian free group.

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