this problem is taken from the herstein book topics in algebra:
G is a finite group such that n divides o(G) define the set H={x/ x^n=e} (is not always a group), prove that the number of elements in H is a multiple of n.
o(G) denotes the number of elements in G.
so this is what i've done so far:
the problem before this is the special case in which G is abelian and is much easier, if G is abelian H is subgroup of G, then if divides n then it divide s o(G) and by Sylow theorem there exist a subgroup S of order in particular
implies and , so and by lagrange theorem divides n, do this for each p prime divisor of n and we get n divides o(H).
as for the general case all i could do was the following: if p divides n and p divides o(G) then p divides o(H).
the key is to observe that H is closed under conjugacy, this is to say if then ,so take some fixed element a of order p (existence asserted by cauchy theorem) and break up H by conjugacy classes relative to (a) (the generated by a), this is to say iff for some ,now the same argument as in the class equation where
N(h)={x/xh=hx} this is because iff . Now take the sum over h in distinct equivalence classes and take mod p, all nontrivial classes vanishes so that , if N(a) is not G then by induction over the size of G we have what we seek.
if N(a) is G then more work is needed ( i know), H is closed under right translation relative to (a) this is to say implies for , this is because every element conmutes with a so that , now break up H by left translation relative to (a) this is iff this gives as in lagrange theorem p distinct equivalence classes with the same number of elements so that p divides o(H).
I hope someone reads this and manage to give a complete solution, any help is welcome.