救命啊。。Abstract Algebra需要帮忙O。。在八月22号前要交哦。。

cutiepipie

不好意思噢。。这些问题必须要用英文来写噢。。请大家帮帮忙。。谢谢。。越快越好。。
Abstract algebra..
1. Let G be a group with |G|= pq, where p>q are distinct primes.
  (i) Show that G has normal Sylow p-subgroup.   
  (ii) If G is non- abelian, then show that q | (p-1) and calculate the number of Sylow q-subgroups in G.
2.Let G be a group with |G|=p²q, where p and q are distinct primes,show that G has a normal Sylow p-subgroups or a normal Sylow q- subgroup.
3.Let G be a group with |G|=p³q,where p and q are distinct primes. Show that either G has a normal Sylow p- or q- subgroup or |G|=24.
4.Let G be a group with |G|=p²q² where p>q are primes. If G does not have a unique normal Sylow p-subgroup, then determine the possible value(s) of |G|.
5.If |G|=pqr where p,q,r are primes, show whether G is simple.
6.If |G|<100 and G is non-abelian and simple, then show that|G|=60.

[ 本帖最后由 cutiepipie 于 21-8-2008 03:20 PM 编辑 ]

dunwan2tellu

1. Let G be a group with |G|= pq, where p>q are distinct primes.
  (i) Show that G has normal Sylow p-subgroup.   
  (ii) If G is non- abelian, then show that q | (p-1) and calculate the number of Sylow q-subgroups in G.

(i) 我们需要一个theorem #:
if P is the only Sylow p-subgroup of G,then P is a normal subgroup of G
设 n_p 为 number of Sylow p-subgroup of G , 那么by Sylow 2 and 3 theorem, n_p = kp+1 for some non-negative integer , k 以及 n_p | pq .
若 k>= 1 ,那么 n_p=kp+1>p>q ， n_p 就不能 divide pq . 所以 k = 0
因此 n_p = 1 . So G contains only one Sylow p-subgroup of order p. 运用 theorem # 我们能总结那个 Sylow p-subgroup is normal .

(ii) From Sylow theorem : n_q = kq + 1 | p , k>= 0 . 所以 kq + 1 = 1 or p . if kq + 1 = 1 then G has unique Sylow q- and unique Sylow p- .任何两个Sylow subgroup 的intersection 之可能是 e (identty element) . 所以 G = Cp x Cq = cyclic group = Abelian . 由于G is non-abelan, 所以 n_q =/= 1
。那么一定是 n_q = p ==> kq + 1 = p ===> q | (p-1) .

2.Let G be a group with |G|=p&sup2;q, where p and q are distinct primes,show that G has a normal Sylow p-subgroups or a normal Sylow q- subgroup

if p>q , then the number of Sylow p-subgroup of G s n_p = kp + 1 | q ==> k=0 . So n_p = 1 and it is normal subgroup by theorem #.



if q>p , then number of Sylow q-subgroup of G n_q = mq+1|p^2 . n_q 只可能是 1,p,p^2 .
case (1) : n_q = 1
那么 theorem # 告诉我们它是 normal .

case(2): if n_q = p
那么 mq + 1 = p 不可能发生，因为 mq+1>q>p  for m>0

case (3) : n_q = p^2
那么 mq + 1 = p^2 ==> mq = (p+1)(p-1) 那么必定 q|(p+1) 因为 q>p>p-1 所以 q 只能整除 p+1
但是如果 q|p+1 表示 p+1 >= q > p .  ==> q=p+1 ==> q=3 , p=2 .

这个 case 里either n_2 = 1 or 3 and n_3 = 1 or 4
proof : if either n_2 =1 or n_3 = 1 , we are done . Else if  n_3 =/= 1 then n_3 = 4 from n_q |p^2 and n_p | q .
当 n_3 = 4 , G 里就有4个不同的 subgroup of order 3,它们之间的共同 element 只有 identity element . 所以除了identity element 外就是那 2x4 = 8 个 element in Sylow 3-subgroup. 也就是说剩下的 |G| - 8 = 12 - 8 = 4 个element 是 in Sylow 2-subgroup of order 4.这正好表示那4个 element 都是在同一个 Sylow 2-subgroup 里，所以他只有 1 个 sylow 2-subgroup .用 theorem # 得知他是 normal subgroup.

3.Let G be a group with |G|=p&sup3;q,where p and q are distinct primes. Show that either G has a normal Sylow p- or q- subgroup or |G|=24

大致上和 Q2 相同方法 prove . 也是分 case q<p 和 q>p 然后一个一个看 n_q = 1,p,p^2,p^3 的时候。发现到 n_q = p^2 时候，|G| = 24

[ 本帖最后由 dunwan2tellu 于 21-8-2008 01:28 AM 编辑 ]

dunwan2tellu

4.Let G be a group with |G|=p&sup2;q&sup2; where p>q are primes. If G does not have a unique normal Sylow p-subgroup, then determine the possible value(s) of |G|.

不清楚你要问什么，因为 |G| = p^2 * q^2 , 还需要 determine ？可能是determine possible number of Sylow p- and q- subgroup ?

5.If |G|=pqr where p,q,r are primes, show whether G is simple.

case(i) : p=q=r then |G|=p^3 , then n_p = kp+1 | p^3 ==> n_p = 1 ==> Sylow p- is normal

case(ii) : p=q =/= r then by Q2 there exist non-trival normal subgroup of G , so G is not simple .

case (iii) : p,q,r distinct .
Let p<q<r . if n_r =1 or n_q = 1 , then there exist normal subgroup of order r or order q , so we are done . Else if n_r and n_q=/= 1 , Then n_r = rk+1|pq ==> n_r = p,q,pq . 但是 r>q>p 所以之可能是 n_r = pq . 同样的 n_q = mq + 1|pr ==> n_q = p,r,pr . 但是 q>p , 所以 n_q=r,pr >= r
所以 the non identity element in Sylow r- = pq(r-1) ; non identty element n Sylow q- is at least = r(q-1) .
Total non identity element in G is at least pq(r-1) + r(q-1) = pqr + r(q-1) - pq > pqr (因为 r>q , q -1 >= p 所以 r(q-1) > pq )
Contradict !所以 n_q or n_r = 1

cutiepipie

那个是要找出当G没有unique normal Sylow p-subgroup 的时候 G 的value 会是什么？
请问哦。。第六题应该怎样做呢？

[ 本帖最后由 cutiepipie 于 21-8-2008 03:07 PM 编辑 ]

dunwan2tellu

原帖由 cutiepipie 于 21-8-2008 03:06 PM 发表
那个是要找出当G没有unique normal Sylow p-subgroup 的时候 G 的value 会是什么？
请问哦。。第六题应该怎样做呢？

题目怎么看都好像怪怪的。我只是猜测他要的是，if G does not hv unique Sylow p-subgroup , then find the values of number of Sylow p-subgroup .

from sylow theorem , number of sylow p -subgroup is n_p = p*k+1 where k>=0 is integer .

since n_p =/= 1 and n_p | q^2 , 那么 n_p 只可能是 q,q^2
但是 p>q 所以只可能是 n_p = q^2

至于第6题，我想应该是要一个个 check case 运用 Question 1,2,3,5 来看每一个所以每一个 |G| in the form of
(i) pq    (ii)p^2q    (iii)p^3q   (iv)pqr  , where p,q,r = prime not 不一定 distinct
must be not simple .

剩下的就是 |G| =
36,48,60,72,80,84,90,96

接下来就是 check 这9个case中，除了 |G|=60 , 其它都是 not simple .
比如
case : |G| = 36 = 2^2 * 3^2
then n_3 = 3k + 1 | 4  . If n_3 = 1 , 那么它就有一个normal subgroup,else if n_3 = 4 , since |G| 不能整除 4! , 那么它有一个normal subgroup
这里我用一个 theorem 和一个 fact
Theorem :
Let G be a finite group and H be its subgroup .Define i(H)=|G|/|H| to be the index of H. if |G| cannot divide (i(H))! , then H must contain a nontrval normal subgroup of G.

Fact :
Let P be the Sylow p-subgroup of G , then  n_p = i(N(P)) = index of N(P)      where N(P) = normaliser of P in G

[ 本帖最后由 dunwan2tellu 于 22-8-2008 02:21 PM 编辑 ]

cutiepipie

第六题的， 我还是不明白咯。。那个fact and theorem 用来做什么哦？ if it is normal subgroups means it not simple ah?

dunwan2tellu

原帖由 cutiepipie 于 21-8-2008 10:41 PM 发表
第六题的， 我还是不明白咯。。那个fact and theorem 用来做什么哦？ if it is normal subgroups means it not simple ah?

为了让你知道我为什么可以从 句子 A 就可以结论 句子 B

if a group has non-trivial normal subgroup , then it is not simple

Q6 也可以参考这里
click这里

[ 本帖最后由 dunwan2tellu 于 22-8-2008 02:38 PM 编辑 ]