|
|
> >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).
yes, since <xy> = {x^i.y^j | i , j in Z},
so, <xy> = {e.e, e.x, e.x^2, ....e.x^(p-1)}U
{y.e, y.x, y.x^2,......y.x^(p-1)}.
let A = {e.e, e.x, e.x^2, ....e.x^(p-1)}
let B = {y.e, y.x, y.x^2,......y.x^(p-1)}
A /\ B = empty.
because,
if e.x^i = y.x^j and i <= j,
then e = y.x^(j-i).
so, y = y^-1 = x^(j-i).
since y not in <x>, contradiction.
thus, |<xy>| = 2p.
> >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.
um...i think...
since |G| = 2*p,
the Sylow 2-subgroups of G are precisely those subgroup of order 2.
and
if P and Q are the Sylow 2-subgroup, gPg^-1 = Q for some g in G.
if P = <x>, P is normal.
so, P = Q.
so, it is the ONLY subgroup of order 2.
is this right ?
so, i need your advice.
> >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.
yes, <x> is the ONLY subgroup of order p.
|
|