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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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Maybe I'm misunderstanding you, but: you won't be able to show that $H$ is outside of $G$, because groups are closed under multiplication. And inverses won't be the problem, since $(g^2)^{-1} = (g^{-1})^2$. So I think that thinking about whether the product of two squares has to be a square (note that this is true if $G$ is abelian!), as The Chaz says, is the thing to do. – Dylan Moreland May 1 '12 at 5:21
For extra points, find a group of order $2^k$ for some $k$ in which the set $H$ is not a subgroup. – Arturo Magidin May 1 '12 at 5:45
I want to prove that normality of H.Let G be a group. Prove or disprove that H={g2|g∈G} is a NORMAL subgroup of G.Can you help me ?? – user35524 Jul 10 '12 at 15:59
@gamzekalkan Please do not post new questions as answers. Use the "Ask Question" button to pose questions and/or post comments to related questions and answers. – Bill Dubuque Jul 10 '12 at 16:04
Great question. Note that this is exercise 54 in chapter 5 in Gallian's Contemporary Abstract Algebra. – a student Jan 11 at 2:26
up vote 11 down vote accepted

$A_4$ 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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... or if you count the number of 3-cycles correctly... – Arturo Magidin May 1 '12 at 5:46
Yes, but if $g = [123]$ then $g^2 = [132]$, and if $g \in G$ then $g$ does not contain the cycle [423]: Therefore, I shouldn't be allowed to square any cycle that isn't in $A_4$ I would have thought. Or should I choose my G to be $S_4$? – Goober May 1 '12 at 17:17
@Goober: No, you don't! You are completely misinterpreting the problem. First, it is impossible for you to have an element $g$ in a group, and have $g^2$ not be in the group: it wouldn't be a group in the first place! Second: the problem is: given a group, in this case $A_4$, construct the set $\{g^2\mid g\in G\}$. Then verify that this set is not always a subgroup (in the case of $G=A_4$, verify that the set of squares of elements of $A_4$ is not a subgroup of $A_4$). It will fail to be a subgroup because the set is not closed under products. But it is always contained in $G$! – Arturo Magidin May 1 '12 at 19:42
@Goober: In other words: a subgroup must (i) be a subset; (ii) contain the identity; (iii) be closed under products; and (iv) be closed under inverses. You seem to think that the only way this $H$ can fail to be a subgroup is if (i) fails. But (i) never fails (and neither do (ii) and (iv); verify this!). In fact, the only way in which $H$ can fail to be a subgroup is if (iii) fails, so in this case you need to verify that (iii) fails. – Arturo Magidin May 1 '12 at 19:58
@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 $A_4$. – Tara B May 1 '12 at 23:57

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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How do you show $x^2y^2$ is not a square? You showed that $(xy)^2\neq x^2y^2$,i.e., $x^2y^2$ is not a square of $xy$. – wxu May 1 '12 at 7:07
@wxu: It's not hard to prove. Why don't you give it a try? Or if you're not very familiar with free groups, then Dylan's answer is probably more suitable for you. – Tara B May 1 '12 at 9:31
@tara, You are right. It is easy. – wxu May 1 '12 at 10:30
@wxu: =] Also, sorry, I assumed you were the person who asked the question, but I see you were not. – Tara B May 1 '12 at 10:39

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
  for i in [1..NrSmallGroups(n)] do
    if(Size(H)<>Size(S)) then count:=count+1; fi;
  if(count>0) then Print(n," ",count,"\n"); fi;
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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 is very neat! – user1729 May 2 '12 at 11:02

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? – a student Jan 11 at 2:25

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