A.3 · Nhóm, lớp coset và ánh xạ thương
GRE Subject Test · GRE Mathematics Chủ đề 9 2:55 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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This unit verifies group structure and computes element orders and subgroup indices; uses kernels and images to identify quotient groups; distinguishes normal subgroups from arbitrary ones; and classifies permutation conjugacy by cycle type.
本单元验证群结构，计算元素的阶和子群的指数；用核与像识别商群；区分正规子群与一般子群；并按轮换型对置换的共轭进行分类。
The order of an element is the least positive power giving the identity; in Z mod n, residue a has order n over the gcd of a and n.
元素的阶是得到单位元的最小正幂；在 Z/nZ 中，剩余 a 的阶是 n/gcd(a, n)。
Lagrange's theorem says subgroup orders divide the group order.
拉格朗日定理说子群的阶整除群的阶。
The converse is not a general existence theorem.
其逆命题不是一般的存在性定理。
Cosets are translates of a subgroup: they have equal size and partition the group, so the index is G over H.
陪集是子群的平移：大小相同并划分整个群，所以指数是 |G|/|H|。
A homomorphism's kernel is always normal, and the first isomorphism theorem identifies G over the kernel with the image, not with the whole codomain.
同态的核总是正规的，第一同构定理把 G/ker φ 与像等同，而不是与整个陪域等同。
Quotient multiplication is well defined only when H is normal.
只有当 H 正规时，商群的乘法才是良定义的。
Conjugation relabels the elements in a permutation's cycles, so conjugacy classes in S n match partitions of n.
共轭只是重新标记置换轮换中的元素，所以 Sₙ 的共轭类对应 n 的划分。
S4 has five classes; 2+1+1 and 2+2 both have order two but are not conjugate.
S₄ 有五个共轭类；2+1+1 和 2+2 的阶都是 2，但不共轭。
Define phi by reducing residues modulo 3.
定义 φ 为把剩余模 3 约化。
It is onto and preserves addition.
它是满射并保持加法。
Its kernel H is 0, 3, 6, 9, so the index is 12 over 4, which is 3.
核 H 是 {0, 3, 6, 9}，所以指数是 12/4 = 3。
The other cosets are 1, 4, 7, 10 and 2, 5, 8, 11.
另外两个陪集是 {1, 4, 7, 10} 和 {2, 5, 8, 11}。
The element 3 has order 4, not 3.
元素 3 的阶是 4，不是 3。
A quotient has one element per coset, not one per element of its kernel.
商群的每个元素对应一个陪集，而不是对应核的每个元素。
And a homomorphism need not be onto its stated codomain.
同态也不一定是到其陪域的满射。
H is 0, 6, 12, so 6 has order three.
H = {0, 6, 12}，所以 6 的阶是 3。
The six cosets r plus H, for r from 0 to 5, partition all 18 residues.
六个陪集 r + H（r 从 0 到 5）划分全部 18 个剩余。
Reduction modulo six is onto with kernel H, so the quotient is Z mod 6, of order six.
模 6 约化是满射，核为 H，所以商群是 Z/6Z，阶为 6。
Normality holds because the group is abelian.
因为群是交换的，所以 H 正规。
Conjugation by 1 2 3 sends 1 2 to 2 3, which is outside H, so H is not normal: cosets still partition S3, but their products depend on representatives.
用 (123) 共轭把 (12) 变成 (23)，它不在 H 中，所以 H 不正规：陪集仍然划分 S₃，但它们的乘积依赖于代表元。
Separately, 1 2 3 times 4 5 has disjoint cycles of length three and two, so its order is 6.
另外，(123)(45) 的不相交轮换长度为 3 和 2，所以它的阶是 6。
Check yourself.
自我检测。
3 has order 4 in Z mod 12.
3 在 Z/12Z 中的阶是 4。
The kernel 0, 3, 6, 9 has index 3.
核 {0, 3, 6, 9} 的指数是 3。
The subgroup e, 1 2 is not normal in S3.
{e, (12)} 在 S₃ 中不正规。
S4 has five conjugacy classes.
S₄ 有五个共轭类。
And 1 2 3 times 4 5 has order 6.
(123)(45) 的阶是 6。