推 teachool :謝謝您的解說 08/29 22:40
※ 引述《teachool (茶)》之銘言：
: 首先感謝各位協助
: 我看了很久不會下筆
: 因為全部混在一起
: 請各位告訴我解題的關鍵
: 先謝謝各位回答
: 1.Find all groups of order 14 up to isomorphism.
[A little Hard]
Let G be order of 14 ＝ 2 ×7. By Sylow theorems,
the Sylow 7-subgroup of G is unique and normal. Set H ＝ ＜ h ＞ and
K ＝ ＜ k ＞ where │H│＝ 7, │K│＝ 2. H is normal in G, so
2
-1 n 2 -2 n -1 n
khk ＝ h . Now, h ＝ k hk ＝ kh k ＝ h , hence
2
n ≡ 1 (mod 7), n ≡ ±1 (mod 7). So
7 2 -1
G ＝ ＜ h,k│h ＝ k ＝ 1, khk ＝ h.＞ ～ Ｚ or
14
7 2 -1 -1
G ＝ ＜ h,k│h ＝ k ＝ 1, khk ＝ h .＞ ～ D .
7
: 2.Find all groups of order 15 up to isomorphism.
[Easy]
Let G be order of 15 ＝ 3 ×5. By 3rd Sylow theorem, both of the
Sylow 3-subgroup and Sylow 5-subgroup are unique and normal.
Thus G ～ ＜ h ＞ ×＜ k ＞ where h and k are of order 3 and 5,
respectively. G ～ Ｚ ×Ｚ ～ Ｚ .
3 5 15
: 我不懂為何出14又出15共兩題
: 差別所在？能否告訴我大致步驟？
數字不同，處理的手法也不同
: 3.Show that any group of order 15 is cyclic.
By Problem 2.
: 4.Prove that any group of order 5 is abelian.
By Lagrange theorem, any group of prime order is cyclic,
and so abelian.
: 我想問15的循環是找15次方會回到1嗎？不懂
對.
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