What are some proofs involving even group orders and subgroup generation in S_n?

[boombaby]

Homework Statement

1, If Card G=2n, which is even, prove that there exists a g in G, such that g^2=e (e=identity)

2, show that Sn=< (1 2), (1 2 3 4...n) >

Homework Equations

The Attempt at a Solution

1, Well, I cannot come up with a practical idea...I've no idea how to find it directly so I assume that there is no g such that g^(-1)=g, then I came to a contradiction, well, in a particular group G with Card G=4...I've no idea how to generalize this procedure to an arbitrary group.

2, I have proved that Sn=< (1 2) (1 3) (1 4)...(1 n) > So I want to show (1 i) can be represented by (1 2) and (1 2 ...n), which will be sufficient. But I checked in S4, and found that the product might be quite difficult to find. (e.g. Let a=(1 2 3 4), b=(1 2) in S4, and (1 3) = abaabaa, (1 4)= babaa )

I'm just start learning group, any suggestion would be greatly appreciated!

 

[morphism]

Hints:

1. Pair off elements with their inverses.

2. What happens when you conjugate an element by a cycle (where conjugating h by g means looking at g^-1 h g)?

 

[boombaby]

1. It turns out to be quite easy.
2. Sorry but I've no idea of what happens really... I checked some different conjugations and was confused, since their results seem quite different and have nothing in common. Could you give me more hints about this one? Thanks

 

[morphism]

Here's an example (I'm going to use 'conjugate' to mean look at ghg^-1 instead of g^-1hg). Let's conjugate (1 2)(3 4) by (1 2 3) in S_4:
(1 2 3)(1 2)(3 4)(3 2 1) = (3 2)(1 4)

Now, the 3-cycle (1 2 3) is the permutation that sends 1 to 2, 2 to 3, 3 to 1, and 4 to 4. What happens when we take the element (1 2)(3 4) and replace each number in it by its image under the permutation (1 2 3)? 1 goes to 2 and 2 goes to 3, so (1 2) becomes (2 3). Similarly, (3 4) becomes (1 4). Thus, (1 2)(3 4) becomes (2 3)(1 4) = (1 2 3)(1 2)(3 4)(3 2 1).

This sort of behavior is true in general: if $\sigma$ and $\tau$ are arbitrary elements of S_n, then $\tau\sigma\tau^{-1}$ has the cycle structure of $\sigma$, where every entry i in $\sigma$ is replaced by $\tau(i)$.

 

[boombaby]

Yea, I get it now. This is pretty cool.

If a(i) = j, then bab^-1( b(i) ) = b(j)
hence they have same structure.

so if I have a=(1 2) and b=(1 2 3 ...n) in hand, then bab^-1 = (2 3) = (1 2)(1 3)(1 2), then we get (1 3). By the same procedure we get (3 4) (4 5)...(n-1 n) = (1 n-1)(1 n)(1 n-1), we get (1 4) (1 5)...(1 n).
Hence <(1 2) (1 2 3...n)>=<(1 2) (1 3)...( 1 n)>=Sn

Thanks!