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 $G$.
 A: 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;

A: 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.
A: There have been some very good counterexamples. Here's another idea to consider. An easy exercice has you prove that a group where all elements square to the identity element is commutative. Also, the set $S$ of all squares is "normal"in $G$ in the sense that for all $g$ in $G$, $g^{-1}Sg=S$.
One direction to look at is simple non commutative groups $G$ : the set of all squares is "normal" and strictly larger than $\{1\}$, thus, if it were a subgroup, it would have to be all of $G$. 
You can then look at the simple noncommutative group $A_6$ and the even permutation $(12)(3456)$. This is the square of no permutation (even in $S_6$). Thus the set of squares of $A_6$ cannot be a subgroup.
A: You can see that if $H$ is a subgroup, then $H$ is also a normal subgroup. Thus a natural way to search for counterexamples is to look at groups with few normal subgroups.
Let $G$ be a finite, nonabelian simple group. Then if $H$ is a subgroup, it must equal $G$, because $g^2 = 1$ for every $g \in G$ implies that $G$ is abelian. If there is an element of order $2$ in $G$, then the map $g \mapsto g^2$ is not injective, thus not surjective, and thus $H$ cannot equal all of $G$.
By Cauchy's theorem, this shows that any finite nonabelian simple group of even order works. So for example, you could pick $G = A_n$ for any $n \geq 5$.
(And by Feit-Thompson, any finite nonabelian simple group works..)
A: 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 $(xy)^2 \not= x^2 y^2$.)
A: $A_4$ 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 $12 = |A_4|$.]
A: Still another approach for finite groups: since squaring is a bijection in odd ordered groups, the set $H$ of squares equals the whole group $G$ if $\mid G \mid$ = odd. It also follows that for groups $G$ of order $2n$, with $n$ odd, the set $H$ is a subgroup indeed. This follows from the fact that in this case the $2$-Sylow subgroup has a normal complement, say $N$ and hence $\mid N \mid$ = $n$ and $|G:N|=2$. Since $N$ consists of squares by the remark in the previous paragraph and all squares must lie in $N$ being of index 2, it follows that $N$ is exactly the set $H$.
