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?
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.
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?