Math Problem Statement

Let Z60 be the group of integers modulo 60 . For each subgroup below, calculate the index of this subgroup:

the subgroup ⟨17⟩ has index [Z60:⟨17⟩]=

Solution

To calculate the index of the subgroup 17\langle 17 \rangle in Z60\mathbb{Z}_{60}, we need to find the number of cosets of 17\langle 17 \rangle in Z60\mathbb{Z}_{60}.

The index of a subgroup a\langle a \rangle in Zn\mathbb{Z}_n is given by the formula:

[Zn:a]=nord(a)[\mathbb{Z}_n : \langle a \rangle] = \frac{n}{\text{ord}(a)}

where ord(a)\text{ord}(a) is the order of the element aa in Zn\mathbb{Z}_n, which is the smallest positive integer kk such that ak0modna^k \equiv 0 \mod n.

Step-by-Step:

  1. Find the order of 17 in Z60\mathbb{Z}_{60}: The order of 17 is the smallest kk such that: 17k0mod6017k \equiv 0 \mod 60 This implies that 17k17k is divisible by 60. To find the order, we calculate the greatest common divisor (gcd) of 17 and 60. Since 17 is a prime number and does not divide 60, we can use the fact that the order of an element in Zn\mathbb{Z}_n is the least number kk such that 17k0mod6017k \equiv 0 \mod 60, or equivalently, 17k17k must be a multiple of 60.

    The order of 1717 in Z60\mathbb{Z}_{60} is the least kk such that: gcd(17,60)=1\gcd(17, 60) = 1

    So, the order of 1717 is (60, this number helps

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Math Problem Analysis

Mathematical Concepts

Group Theory
Modular Arithmetic
Subgroups
Index of a Subgroup

Formulas

[Z_n : ⟨a⟩] = n / ord(a)
ord(a) is the smallest k such that a * k ≡ 0 (mod n)

Theorems

Lagrange’s Theorem

Suitable Grade Level

Undergraduate Mathematics