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In article <eeb1d6$ka8$1@xxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> 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.)
Yes. (Or by Sylow's Theorem).
>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.)
If G has a normal subgroup of order 2, generated by y, then
<y>/\<x>={1}; and since both are normal, you have that
[y,x] = y^{-1}x^{-1}yx = y^{-1}(x^{-1}yx) lies in <y> and
[y,x] = y^{-1}x^{-1}yx = (y^{-1}(x^{-1}y)x lies in <x>
(by normality). So [y,x] = 1. Therefore,
y^{-1}x^{-1}yx = 1
or
yx = xy.
So x and y commute.
Therefore, the order of xy is 2p (since the orders of x and y are
relatively prime, and they commute, the order of xy is the product of
the orders).
>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,
Do you know Sylow's Theorems yet? Since the subgroup of order 2 is
normal, it is the ONLY subgroup of order 2, so this would require
xy=y.
>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,
Again, if xy were of order p, by Sylow you would have that xy lies in
<x>, which would imply that y is a power of x.
>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>.
Indeed.
>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.
See above.
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
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