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Julien Santini wrote:
> Given a group G with order n, p an odd prime, and a_1,...,a_p p
> elements of G such that a_1a_2=a_2a_3=...=a_pa_1, show that n is
> divisible by 2p.
>
> ===
> I could show that n is divisible by 2 (all a_i have same even order),
> (a_i)^2=(a_k)^2 for all i,k, a_ia_k=a_(i+1)a_(k+1) for all i,k. The
> last assertion gives that the set {a_i,a_j; p>=i,j>=1} is a subgroup of
> G with order p, provided that we can prove that (a_i)^2=1 for at least
> one i. Can we show this ?
> ===
>
> Thanks.
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Hi:
Of course: if you already proved that |G| = n is divisible by two, then
apply Cauchy's Theorem
to deduce that there's an element of order 2 in G.
Regards
Tonio
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