Permutation/Alternating Group

[Kalinka35]

Homework Statement

Let Q_n = {x in S_n | x=y² for some y in S_n}.

Is it true or false that Q_n = A_n?

Homework Equations

The Attempt at a Solution

I'm inclined to say it's false because of the strangeness of S₂. Shouldn't A₂ contain both the identity and (12)? But (12)(12) is the identity so it seems there is no way to get (12) by the definition of Q. Is this correct or am I missing something?

 

[latentcorpse]

no S2 has 2!=2 elements, A2 has 2!/2=1 elements (the identity)

 

[Kalinka35]

Oh yes, I can't believe I overlooked that. But I'm still not sure whether Q is equal to the alternating group...

 

[matt grime]

Do you know of any set of elements that generates A_n?

Forget that - that's wrong. I thought you had to look at the group generated by Q_n since it is not even clear that Q_n is a subgroup of S_n, nevermind being equal to A_n.

Instead, have you tried to see if it's true for n=3,4 or 5?

 

[matt grime]

Ok, let me go further: have you checked if it is true for n=6? (perhaps via a computer algebra package).

If you don't have one (a computer algebra package I mean), let's try thinking about why it might be false (rather than finding a counter example).

If g has order k, what is the order of g^2? (There are two cases: k even and k odd.)

 

[Kalinka35]

It seems like it is true from the examples I worked out. For instance, the square of a 5-cycle is a 5-cycle, and in general the square of a cycle of odd length n will produce another cycle of odd length n.
For even n, I am getting that for length 2n you get two n-cycles.
So I am thinking it is probably true. I can't seem to find any reasoning why it wouldn't be at this point.

 

[matt grime]

Well, I'm afraid you're wrong: Q_n isn't even a group in general, nevermind being A_n.

I think you have the hint the wrong way round. Suppose that I have element of order 4 in A_n. What order must its 'square root' have if it were to exist.

 

[Kalinka35]

Okay, here is what I think may be a counterexample:
the permutation (1234)(56) is in A6, but it cannot be produced by squaring any element in S6. So yes, you're right Qn is not equivalent to An.

 

[matt grime]

How do you know it is not the square of any element in S_6? (It isn't, but you need a proof of this fact.)