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【翻译】:
定理:对于每个素数 p,直到同构,恰好有两个 p2 阶的非同构群; 这些是 Za 和 Z ×Zp。
证明:根据定理 5.11,Z,a 恰好有一个 p 阶子群,而 Zp×Zp 至少有两个:Zpx{0} 和 {0}×Zp。 因此这两组是非同构的。
假设 G 有阶 p² 并且不与 Zp² 同构。 这意味着 G 不是循环的。 根据柯西定理 11.3 有 a ∈ G 且 ord a =p。对于 b ∉ (a) 我们有 ordb = por = p2; 后者是不可能的,因为这意味着 G 是循环的。 所以 ord b = p 和 I(b)I= p。 注意,由于 b ∉ (a) 和任何素数阶群都不能有非平凡子群,我们有 (a)∩(b)={e}。 显然 Ka)(b)/=KalI(b)l= p= |G| 因此 G =(a)(b)。 由 12.1 G 中的第一次观察,G 是阿贝尔的,因此 (a) 和 (b) 是正规子群,因此定理 8.5 和定理 9.12 表明 G≌Z,×Zp。
【原文】
Theorem. For every prime p there are, up to isomorphism, precisely two non-isomorphic groups of order p2; these are Za and Z ×Zp.
Proof. By Theorem 5.11,Z,a has precisely one subgroup of order p, whereasZp×Zp has at least two: Zpx{0} and {0}×Zp. The two groups are therefore non-isomorphic.
Assume G has order p² and is not isomorphic to Zp². This means G is not cyclic. By Cauchy's Theorem 11.3 there is a ∈ G with ord a =p.For b ∉ (a) we have ordb = por = p2; the latter is impossible since this would imply G is cyclic. So ord b = p and I(b)I= p. Note that, since b ∉ (a) and any group of prime order cannot have non-trivial subgroups, we have(a)∩(b)={e}. Clearly Ka)(b)/=KalI(b)l= p= |G| and thus G =(a)(b). By the first observation in 12.1 G is abelian, hence (a) and (b) arenormal subgroups, and so Theorem 8.5 and Theorem 9.12 show thatG≌Z,×Zp.
定理:对于每个素数 p,直到同构,恰好有两个 p2 阶的非同构群; 这些是 Za 和 Z ×Zp。
证明:根据定理 5.11,Z,a 恰好有一个 p 阶子群,而 Zp×Zp 至少有两个:Zpx{0} 和 {0}×Zp。 因此这两组是非同构的。
假设 G 有阶 p² 并且不与 Zp² 同构。 这意味着 G 不是循环的。 根据柯西定理 11.3 有 a ∈ G 且 ord a =p。对于 b ∉ (a) 我们有 ordb = por = p2; 后者是不可能的,因为这意味着 G 是循环的。 所以 ord b = p 和 I(b)I= p。 注意,由于 b ∉ (a) 和任何素数阶群都不能有非平凡子群,我们有 (a)∩(b)={e}。 显然 Ka)(b)/=KalI(b)l= p= |G| 因此 G =(a)(b)。 由 12.1 G 中的第一次观察,G 是阿贝尔的,因此 (a) 和 (b) 是正规子群,因此定理 8.5 和定理 9.12 表明 G≌Z,×Zp。
【原文】
Theorem. For every prime p there are, up to isomorphism, precisely two non-isomorphic groups of order p2; these are Za and Z ×Zp.
Proof. By Theorem 5.11,Z,a has precisely one subgroup of order p, whereasZp×Zp has at least two: Zpx{0} and {0}×Zp. The two groups are therefore non-isomorphic.
Assume G has order p² and is not isomorphic to Zp². This means G is not cyclic. By Cauchy's Theorem 11.3 there is a ∈ G with ord a =p.For b ∉ (a) we have ordb = por = p2; the latter is impossible since this would imply G is cyclic. So ord b = p and I(b)I= p. Note that, since b ∉ (a) and any group of prime order cannot have non-trivial subgroups, we have(a)∩(b)={e}. Clearly Ka)(b)/=KalI(b)l= p= |G| and thus G =(a)(b). By the first observation in 12.1 G is abelian, hence (a) and (b) arenormal subgroups, and so Theorem 8.5 and Theorem 9.12 show thatG≌Z,×Zp.
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