Thread: Group Theory

I have the following problem which i have no idea how to proceed.

Let G be a finite group with odd number of elements. Show that for every , there exists an element such that .

I have tried to construct a subgroup H of G defined by However, I cant proceed on after that. Perhaps someone can guide me on how to proceed on with this problem. Thank You.

hint: use lagrange's theorem, to show that the order of every element of G is odd. now show that <g> = <g^2> for every g in G.

why can we use this to find h?

let me try the solution with the help of ur hint.

Let K = <g>. Since K is cyclic, the order of K = order of g. Also, K is a subgroup of G. By lagrange theorem, divides . Since is odd, therefore is also odd. Hence, order of any elements in G is odd.

Next to show: < > = < >.
For this part i am not very sure but i will try:
<g> = { , ...} and < > = { , ...}. Since order of g is odd, therefore, g^(2m+1) = 1, where 1 is the identity of G. Hence, <g> reduces to <g> = { ...}. Thus <g> = <g^2>.

Is the above reasoning correct?
However I dont really know how this allows us to find h.

for the second part, since the powers of g^2 "go up by 2's", and since |G| = 2m+1, when we get to g^(2m+2), we wind up back at g.

now surely <g^2> is a subgroup of <g> (every even power of g is also a power of g).

but since g is in <g^2>, <g> must be a subgroup of <g^2>, so.... (fill in the blank).

but this means that g = g^(2m+2) = [g^(m+1)]^2, so what do you think we should choose for h?

Choose h= g^(m+1) ?

and does h^2 = g then? if so, you're done.