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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.
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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.)
since G has a normal subgroup of order 2,
let N = {e, y} be normal subgroup of G.
(of course, y^2 = e and y^-1 = y.)
i will show that |xy| = 2p.
by Lagrange, |xy| | 2p.
so, |xy| = 1 or 2 or p or 2p.
1) if |xy| = 1,
then xy = e => y^-1 = x => x = y (because, y^-1 = y)
since |<x>| = p and |<y>| = 2, contradiction.
2) if |xy| = 2,
then xyxy = e => xy = y^-1.x^-1 => xy = y.x^-1
(because, y^-1 = y).
and
since N is normal, xyN = Nxy.
so, {xy, xyy} ={xy, yxy}.
so, xyy = yxy => xy = yx.
so, xy = y.x^-1 = yx.
so, x^-1 = x.
so, |<x>| = 2. contradiction.
3) if |xy| = p,
then (xy)^p = e.
since <x> is normal,
<x> = <x>.(xy)^p = (<x>xy)^p = (<x>y)^p = <x>.y^p
since p is odd and y^2 = y,
<x>.y^p = <x>y.
so, <x> = <x>y.
so, y in <x>.
so, |<y>| | <x>.
so, 2 | p. contradcition.
thus, |xy| = 2p.
thus, <xy> = G. so, cyclic.
is this right ??
if you have other nice solution, i hope to know that.
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