The set $\{g^2 | g \in G\}$ in a group $G$

Let $G$ be a group. Prove or disprove that $H =\{g^2 | g \in G\}$ is a subgroup of $G$.

I tried testing the permutations of $A_4$, however squaring each cycle yielded a cycle in $A_4$ so I'm lacking a counter-example (if there is one). In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup.

Or, I could be way off base and figure out that there isn't a counter-example and I need to prove that indeed $H$ is a subgroup of $G$.

Solution 1:

As you suspected, the statement is false. Consider the free group $G$ on two generators, say $x$ and $y$. Then $x^2$ and $y^2$ are both in $H$, but there is no way to write $x^2 y^2$ as a square. (Remember, $x$ and $y$ don't commute, so $(xy)^2 \not= x^2 y^2$.)

Solution 2:

$A_4$ should work. The squares will lie in $A_4$ simply because a group is closed under its multiplication, but there are further obstructions to a subset being a subgroup.

Now, what are the squares in $A_4$? We have two types of non-trivial elements in $A_4$: products of disjoint transpositions such as $[12][34]$, which square to the identity, and $3$-cycles such as $\sigma = [123]$. The $3$-cycles satisfy $\sigma^3 = e$ and hence $\sigma = (\sigma^{-1})^2$, so they are all squares.

So, does the set of all $3$-cycles, together with the identity, form a subgroup of $A_4$? If I've multiplied $[123][423]$ out correctly, then the answer is no. [Another reason, which I think is what Arturo is getting at: there are $8$ different $3$-cycles, and $9$ does not divide $12 = |A_4|$.]

Solution 3:

May I offer a computational solution?

If you run this code in GAP, it will give you the number of non-isomorphic groups $G$ of order $n \leq 200$ for which $H$ is not a subgroup. It encounters thousands of counter-examples in a matter of seconds.

for n in [1..200] do
  count:=0;
  for i in [1..NrSmallGroups(n)] do
    G:=SmallGroup(n,i);
    S:=Set(G,g->g^2);
    H:=Group(S);
    if(Size(H)<>Size(S)) then count:=count+1; fi;
  od;
  if(count>0) then Print(n," ",count,"\n"); fi;
od;

Solution 4:

Your idea of what you would have to do to find a counterexample was confused.

"In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup."

What you are trying to do is prove that there exists a group $G$ such that the set $H$ is not a subgroup. So the counterexample you are looking for is the group $G$, not some subgroup of it. You correctly guessed that $A_4$ would be an example, but then what you need to show is that the set of all squares of elements of $A_4$ doesn't form a subgroup. Clearly the problem is not going to be that the set of all squares is not even contained in $A_4$, because $A_4$ is a group. But you can show (as Dylan has mentioned) that there exist two squares in $A_4$ such that their product is not a square.

I'm not sure why I wrote this, since others have already said as much. I guess I just wasn't quite sure whether you have seen why your approach didn't make sense.