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In article <eebvbf$o5$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).
>
>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.
Oy. And people (rightly) complain that I make my proofs too
complicated...
Let G be a group, let x and y be elements of G such that
xy=yx. Suppose further that the order of x and the order of y are
relatively prime. Then the order of xy is the product of the orders of
x and y.
Proof: Let order(x)=n, order(y)=m. Since x and y commute, (xy)^k = x^k
y^k. Since the orders are relatively prime, <x>/\<y>={1} (the
intersection is a subgroup of both, hence the order must divide both n
and m). As (xy)^{nm} = (x^n)^m*(y^m)^n = 1, it follows that the order
must divide nm. If (xy)^k = 1, then x^k=y^{-k}, so x^k in <x>/\<y> =
{1}, hence m|k. Symmetrically, n|k, so nm|k. Thus, the order is a
multiple of nm, proving order(xy)=nm.
The result can be generalized. In general, one can prove that if
order(x)=m, order(y) = n, and xy=yx, then
lcm(m,n) | lcm(m,n) | |
-------- | --------------- | order(xy) | lcm(m,n)
gcd(m,n) | order(<x>/\<y>) | |
but you hardly need this much.
>> 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 ?
Yes. In general, a p-Sylow subgroup is normal if and only if it is the
ONLY subgroup of that order.
>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.
Exactly.
--
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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)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
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