Nhóm, lớp đồng dư và ánh xạ thương
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| coset/ˈkɒset/ | Lớp coset |
| normal subgroup/ˈnɔːml ˈsʌbɡruːp/ | Nhóm con chuẩn |
| index/ˈɪndeks/ | A number indicating the power to which a base is raised. |
Grouping the clock
- Twelve clock hours can be sorted into three classes — $\{0,3,6,9\}$, $\{1,4,7,10\}$, $\{2,5,8,11\}$ — and adding hours still works class by class.
- That is a quotient group. GRE group theory tests orders, cosets 陪集, Lagrange, kernels and normal subgroups 正规子群.
Orders and Lagrange
- The order of $g$ is the least $k > 0$ with $g^k = e$. In $\mathbb Z/n\mathbb Z$, the order of $a$ is $\dfrac{n}{\gcd(a,n)}$: in $\mathbb Z/12$, the element 3 has order 4.
- Lagrange: subgroup orders divide $|G|$. The converse fails in general: $A_4$ (order 12) has no subgroup of order 6.
What is the order of 8 in $\mathbb Z/20\mathbb Z$?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
In ℤ/nℤ the order of a is n / gcd(a, n) = 20 / gcd(8, 20) = 20 / 4 = 5. Check: 5 · 8 = 40 ≡ 0.
A group of order 12 need not have a subgroup of every order dividing 12.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. The converse of Lagrange's theorem fails: $A_4$ has order 12 but no subgroup of order 6.
Cosets and index 指数
- $g + H$ is a translate of $H$. Cosets have equal size and partition $G$, so the index is $|G|/|H|$.
- $a, b$ share a coset exactly when $a - b \in H$. A coset other than $H$ is not a subgroup: it misses the identity.
Which coset of H = {0, 3, 6, 9}?
Place elements of ℤ/12 in cosets by their difference test.
A subgroup of order 6 sits inside a group of order 42. What is its index?
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
By Lagrange's theorem the index is |G| / |H| = 42 / 6 = 7, the number of cosets.
Kernels and quotients
- $\ker\varphi$ is normal and $G/\ker\varphi \cong \operatorname{im}\varphi$ — the image, which equals the codomain only if $\varphi$ is onto.
Reduce mod 3: $\varphi: \mathbb Z/12 \to \mathbb Z/3$. Onto, with kernel $H = \{0,3,6,9\}$, $|H| = 4$, index 3. So $(\mathbb Z/12)/H \cong \mathbb Z/3$: one element per coset.
Normality and permutations
- $G/H$ is a group only if $H$ is normal: $gHg^{-1} = H$. Abelian subgroups and index-2 subgroups always are.
- In $S_n$: order = lcm of disjoint cycle lengths; $(1\,2)(3\,4\,5)$ has order 6. Conjugate permutations have the same cycle type, so $S_4$ has 5 conjugacy classes (partitions of 4).
What is the order of $(1\,2\,3)(4\,5\,6\,7)$ in $S_7$?
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The cycles are disjoint, so the order is the lcm of their lengths: lcm(3, 4) = 12.
How many conjugacy classes does $S_4$ have?
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Conjugacy classes in Sₙ correspond to cycle types, i.e. partitions of n. The partitions of 4 are 4, 3+1, 2+2, 2+1+1 and 1+1+1+1, so there are 5 classes.
Recap
- Order of $a$ in $\mathbb Z/n$: $n/\gcd(a,n)$. Lagrange has no general converse.
- Index = number of cosets; $G/\ker \cong$ image.
- Quotients need normal subgroups; cycle type determines conjugacy in $S_n$.