
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 $\displaystyle g \in G$, there exists an element $\displaystyle h \in G$ such that $\displaystyle g=h^2$.
I have tried to construct a subgroup H of G defined by $\displaystyle H=\{h\in G \mid g=h^2, \forall g \in G \}. $ However, I cant proceed on after that. Perhaps someone can guide me on how to proceed on with this problem. Thank You.

Re: Group Theory
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?

Re: Group Theory
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, $\displaystyle \left\vert K \right\vert$ divides $\displaystyle \left\vert G \right\vert$. Since $\displaystyle \left\vert G \right\vert$ is odd, therefore $\displaystyle \left\vert K \right\vert$ is also odd. Hence, order of any elements in G is odd.
Next to show: <$\displaystyle g$> = <$\displaystyle g^2$>.
For this part i am not very sure but i will try:
<g> = {$\displaystyle 1, g, g^2, g^3, g^4$, ...} and <$\displaystyle g^2$> = {$\displaystyle 1, g^2, g^4, g^6$, ...}. Since order of g is odd, therefore, g^(2m+1) = 1, where 1 is the identity of G. Hence, <g> reduces to <g> = {$\displaystyle 1, g^2, g^4,$ ...}. Thus <g> = <g^2>.
Is the above reasoning correct?
However I dont really know how this allows us to find h.

Re: Group Theory
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?

Re: Group Theory

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