What is Generated subgroup; order? — Mathematics

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Definition 1.13 University Mathematics — Year 2 · অধ্যায় 1 — Sets and Structures

Let $G$ be a group and $A \subseteq G$ . The subgroup generated by $A$ , written $\langle A \rangle$ , is the smallest subgroup containing $A$ — concretely, all finite products of elements of $A$ and their inverses. A group is cyclic when generated by one element: $\langle a\rangle = \{a^k : k \in \Z\}$ . The order of $a \in G$ is $\operatorname{ord}(a) = \abs{\langle a \rangle}$ (possibly infinite); when finite, it is the least $n \geq 1$ with $a^n = e$ , and $a^k = e \iff \operatorname{ord}(a) \mid k$ .

The subgroup lattice of ℤ/12ℤ: one subgroup per divisor of 12 (), with an edge when one contains the other with prime index. Inclusions run against divisibility of the generator: 4 ⊂eq 2 because 4 is a multiple of 2. — The subgroup lattice of $\Z/12\Z$ : one subgroup per divisor of $12$ (Proposition 1.17), with an edge when one contains the other with prime index. Inclusions run *against* divisibility of the generator: $\langle\overline 4\rangle \subseteq \langle\overline2\rangle$ because $4$ is a multiple of $2$ .

উদাহরণ

Example 1.15 (Cosets in action: $A_3$ inside $\mathfrak{S}_3$ )

Take $G = \mathfrak{S}_3$ (order $6$ ) and $H = A_3 = \{\mathrm{id},\ (1\,2\,3),\ (1\,3\,2)\}$ . The left cosets are

H = \{\mathrm{id},\ (1\,2\,3),\ (1\,3\,2)\}, \qquad (1\,2)H = \{(1\,2),\ (2\,3),\ (1\,3)\} :

two classes of three elements partitioning $G$ , exactly as the count $\abs G = \abs H \times (\text{number of cosets})$ demands — and visibly the partition into even and odd permutations. Note $(1\,3)H = (1\,2)H$ although $(1\,3) \neq (1\,2)$ : cosets are classes, not labelled by their representatives, and $x^{-1}y \in H$ is the only legitimate comparison. This two-class picture is the general one for the signature: $A_n$ and its lone companion coset split $\mathfrak{S}_n$ in half, which is how the weekend problem counts reachable puzzle positions.

Example 1.16

Two immediate dividends. Groups of prime order are cyclic: if $\abs G = p$ is prime and $a \neq e$ , then $\operatorname{ord}(a)$ divides $p$ and is not $1$ , so it is $p$ : $\langle a\rangle = G$ . The subgroup lattice of $\Z/12\Z$ : by Proposition 1.17 below, there is exactly one subgroup per divisor of $12$ — orders $1, 2, 3, 4, 6, 12$ , generated respectively by $\overline 0$ , $\overline 6$ , $\overline 4$ , $\overline 3$ , $\overline 2$ , $\overline 1$ . The closing caution: the converse of Lagrange fails in general — $A_4$ has order $12$ but no subgroup of order $6$ , as we prove in this chapter’s weekend problem (Problem 1.1, question 14). Lagrange restricts the possible orders; it does not promise them.

Example 1.20 (Cycle type as a census)

How many permutations of $\mathfrak{S}_9$ have the cycle type $(4, 3, 2)$ — one $4$ -cycle, one $3$ -cycle, one transposition? Choose the supports and the cyclic orders:

\frac{9!}{4\cdot 3\cdot 2} = \frac{362\,880}{24} = 15\,120 :

list the nine symbols in a row ( $9!$ ways), bracket the first four, next three, last two into cycles, and divide by the rotations inside each bracket ( $4$ , $3$ and $2$ of them) which give the same permutation. (Distinct cycle lengths here, so no further division; equal lengths would also require dividing by the permutations of the equal brackets.) Every such permutation has order $\operatorname{lcm}(4,3,2) = 12$ and signature $(-1)^3(-1)^2(-1)^1 = +1$ (Theorem 1.19 and the signature theorem below). One partition of $9$ , one conjugacy class, one census — the combinatorics of $\mathfrak{S}_n$ is the arithmetic of partitions.

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