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mina_world wrote:
> hello sir~
>
> let G be a group of order 2p where p is an odd prime.
>
> show that if G has a normal subgroup of order 2,
> then G is cyclic.
>
> ----------------------------------------------
> um...i think....
>
> by Cauch's theorem,
> G has an element "x" of order p.
> so, let |<x>| = p.
> (of course, <x> is normal because |<x>|=|G|/2.)
>have other nice solution, i hope to know that.
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Hi:
Wow, what along answer! Simpler: if G has sbgp.of order 2, then since
it surely has
a normal (Sylow) sbgp. of order p (why?), then G is the (inner) direct
product of these
two sbgps., and thus obviously abelian. But this dir. prod. is
isomorphic with the
cyclic group of order 2p (Chinese Remainder Theorem when dealing with
rings, say...)
Regards
Tonio
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