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In article <eeb1d6$ka8$1@xxxxxxxxxxxxxxxx>,
"mina_world" <mina_world@xxxxxxxxxxx> writes:
>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.)
>
>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.
I think your solution below is correct (I have not checked every detail),
but it seems unnecessarily long. It is preferable to use a more general
technique.
Lemma. If normal subgroups M and N of a group G satisfy M intersect N = {e},
then mn = nm for all m in M and n in N.
Proof By normality, the commutator m^-1 n^-1 m n lies in both M and N, hence
m^-1 n^-1 m n = e, so mn = nm.
Now in your problem, N and <x> have coprime orders, so N intersect <x> = {e}.
Hence xy = yx, from which it follows easily that |xy| = 2p
(because (xy)^2 = x^2y^2 = x^2 != e, and (xy)^p = x^py^p = y^p = y != e).
Derek Holt.
>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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