Finite group

[xxxxtt]

We have a finite group G with 4n+2 elements,n>0.
Find out the number of elements x from G such that x^(2n+1)=e

[amfulger]

If there is an element x in the group of order 2n+1, then by Cauchy there is an element y of order 2 and the gropu is described by the relations x ²ⁿ⁺¹=y ² =e and xy=yx. We have 2n+1 elements a such that a ²ⁿ⁺¹=e. These are the elements of the group <x>.
However there may not be an element of order 2n+1.

[grobber]

Here are some more in the same context. Maybe one of them will help:

1. Let G be a group with 4n+2 elements. Prove that there is at most one subgroup with 2n+1 elements. (I have a solution for this one)

2. Let G be a group with 4n+2 elements. Let x and y be distinct elements s.t. x,y=/=e and x^2=y^2=e. Prove that (xy)^(2n+1)=e. (I don't have a soln for this one)

[amfulger]

If G is commutative the by the Structure of Commutative Finite Group Theorem we get G izomorphic to
G'=Z_n1*Z_n2*...*Z_nk for some integer k and n₁*...*n_k=4n+2 and
n₁|...|n_k.
From these we get n_k is even and all the others are odd.
It is easy to see the elements of order 2n+1 in G' are the elements of the form (a₁,...,2*a_k) where a_i is in Z_ni.
By counting these elements we get n₁*n₂*...*(n_k/2)=2n+1.
So if G is commutative there are 2n+1 elements x of such that x ²ⁿ⁺¹=e or in additive notation (2n+1)x=0.

[grobber]

I have the answer to this problem in a book. I ind of cheated ( ) and looed at the last line of the solution. I thin it said there that the answer is <=2n+1, so maybe any number <= 2n+1 of elements for which x^2=e can be found. maybe we could start this one by showing that the number must be <=2n+1?

[grobber]

According to this problem, the answer is .

[alekk]

hi amfulger,

do you have a nice proof of "the Structure of Commutative Finite Group Theorem " because I've once seen a ugly proof in one of my book but i think it must exist a nicer one.

regards.

[grobber]

I saw a reasonable proof in a file, but I forgot where I got the file from and the extension .ps is not allowed for uploading, so I can't put it here .