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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 $G$.

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    Great question. Note that this is exercise 54 in chapter 5 in Gallian's Contemporary Abstract Algebra.2016-01-11

7 Answers 7

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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 $index|G:N|=2$. 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 $H$.

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    Your answer seems to be saying "yes, the set in question is indeed a subgroup". But this contradicts the other answer(s) which provide counter examples. Please could you elaborate?2016-01-11
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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 $(xy)^2 \not= x^2 y^2$.)

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    @wxu: =] Also, sorry, I assumed you were the person who asked the question, but I see you were not.2012-05-01
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$A_4$ 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 $12 = |A_4|$.]

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    @Goober: Did you not read the answers? I think it had been pretty well explained to you already that you don't need, and can't get, $g^2$ not to be in $A_4$.2012-05-01
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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; 
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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.

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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.

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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..)

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    I just thought I'd say that I like this approach very much...it is very neat!2012-05-02