a 發問時間: 科學數學 · 10 年前

structure of abelian p-group

請問為什麼若A是個 abelian p-group

a1 屬於 A, A1 是 a1 展開的cyclic group

則 by induction

A/A1=A2+A3+...+As (direct sum)

他上面沒有說A2...是什麼

已更新項目:

忘了說 a1 has maximal period

1 個解答

評分
  • 10 年前
    最佳解答

    厄~~

    如果他真的寫的這麼少,就真的有點難懂了

    要證明A/A1=A2+...+As是direct sum

    需要一個輔助定理:

    Given any x belong to A/A1, and x has period p^r. There exist a x' belong to A, and x' also has period p^r.

    pf: Given any x belong to A/A1, and has period p^r. There exist a x1 belong to A, such that (p^r)x1=na1

    if n=0, trivial

    if not, write n as (p^k)t, s.t. (p,t)=1 Since t is relative prime to p, ta1 also has period p^r1.

    we have (p^r)x1=(p^k)ta1 imply p^(r+r1-k)x1=0

    notice that (r+r1-k)<=r1, so r<k.

    we have (p^r)x1=(p^r)[p^(k-r)]ta1, notice that [p^(k-r)]ta1,said c, belong to A1.

    If x1 belong to A1, trivial

    If not, there exist (p^r)(x1-c)=0 and (period of x1-c)<p^r

    (p.s. remember x has period p^r)

    choose x'=x1-c, and we have done.

    證完以後就可以用induction了

    最後再補充一下

    如果(A2/A1)和(A3/A1)有非0 交集(這裡A2/A1就是你所說的A2,不要搞混囉)

    讓order of (A2/A1)>= (A3/A1)好了, let (A3/A1) genetated by a3=(a3)'A1

    再用一次上面的lemma, let A/A2 be the factor group, (a3)'A2 有period p^r

    則存在 (a3)''屬於A 有period p^r

    那我們就選一個新的A3/A1, generated by (a3)''A1

    不就是direct sum了嗎?

    by induction 可以一直不斷的做下去, 因為是finite group,總會做完

    它的意思就是降

    2010-01-25 00:05:37 補充:

    他媽的,預覽還是好的,只要有prime就變亂碼 氣死我了

    2010-01-25 00:08:53 補充:

    厄~~

    如果他真的寫的這麼少,就真的有點難懂了

    要證明A/A1=A2+...+As是direct sum

    需要一個輔助定理:

    Given any x belong to A/A1, and x has period p^r. There exist a x' belong to A, and x' also has period p^r.

    pf: Given any x belong to A/A1, and has period p^r. There exist a x1 belong to A, such that (p^r)x1=na1

    if n=0, trivial

    2010-01-25 00:09:49 補充:

    if not, write n as (p^k)t, s.t. (p,t)=1 Since t is relative prime to p, ta1 also has period p^r1.

    we have (p^r)x1=(p^k)ta1 imply p^(r+r1-k)x1=0

    notice that (r+r1-k)<=r1, so r

    2010-01-25 00:10:42 補充:

    notice that (r+r1-k)<=r1, so r

    2010-01-25 00:12:14 補充:

    notice that (r+r1-k)<=r1, so r<=k.

    we have (p^r)x1=(p^r)[p^(k-r)]ta1, notice that [p^(k-r)]ta1,said c, belong to A1.

    If x1 belong to A1, trivial

    If not, there exist (p^r)(x1-c)=0 and (period of x1-c)

    2010-01-25 00:17:51 補充:

    (p.s. remember x has period p^r)

    choose x``;=x1-c, and we have done.

    證完以後就可以用induction了

    最後再補充一下

    如果(A2/A1)和(A3/A1)有非0 交集(這裡A2/A1就是你所說的A2,不要搞混囉)

    讓order of (A2/A1)>= (A3/A1)好了, let (A3/A1) genetated by a3=(a3)``A1

    再用一次上面的lemma, let A/A2 be the factor group, (a3)``A2 有period p^r

    則存在 (a3)``屬於A 有period p^r

    2010-01-25 00:18:26 補充:

    那我們就選一個新的A3/A1, generated by (a3)``A1

    不就是direct sum了嗎?

    by induction 可以一直不斷的做下去, 因為是finite group,總會做完

    它的意思就是降

    2010-01-25 00:19:44 補充:

    不好意思我打了一個多小時

    預覽還是好的

    貼上來就全部都變亂碼

    用補充的變得很亂

    我覺得這是yahoo的問題

    預覽看到的和po出來的應該要一樣

    參考資料: Undergraduate Algebra
還有問題?馬上發問,尋求解答。