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Thread: challenging group theory problem

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    challenging group theory problem

    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 $\displaystyle p^k$ divides n then it divide s o(G) and by Sylow theorem there exist a subgroup S of order $\displaystyle p^k$ in particular
    $\displaystyle a\in S$ implies $\displaystyle a^p^k=e$and $\displaystyle a\in H$, so $\displaystyle S\subset H$ and by lagrange theorem $\displaystyle p^k$ 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 $\displaystyle h\in H$ then $\displaystyle g^-hg\in H$,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 $\displaystyle h\sim g$ iff $\displaystyle b^-hb=g$for some $\displaystyle b\in (a)$,now the same argument as in the class equation $\displaystyle o([h])=\frac{o(a)}{N(h)\cap(a)}$ where
    N(h)={x/xh=hx} this is because $\displaystyle b^-hb=c^-hc$ iff $\displaystyle N(h)\cap(a)b=N(h)\cap(a)c$. Now take the sum over h in distinct equivalence classes and take mod p, all nontrivial classes vanishes so that $\displaystyle o(H)=o(N(a)\cap H) mod p$ , 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 $\displaystyle h\in H$ implies $\displaystyle hb\in H$ for $\displaystyle b\in (a)$, this is because every element conmutes with a so that $\displaystyle (ha)^n=h^na^n$, now break up H by left translation relative to (a) this is $\displaystyle g\sim h$ iff $\displaystyle ga^k=h$ 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.
    Last edited by paragrafo; Apr 4th 2011 at 06:38 PM.
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