Let G be a cyclic group i Find the order of each element ii

Let G = be a cyclic group. (i) Find the order of each element. (ii) Find the set of all left cosets of G by the subgroup H = . (iii) Find the order of each clement of the factor group G/H.

Solution

a) order of 1, is 1

Order of a, is 12; a¹² = 1

Order of a² is 6; (a²)⁶ = 1

Order of a³ is 4; (a³)⁴ = a¹² = 1

Order of a⁴ is 3; (a⁴)³ = a¹² = 1

Order of a⁵ is 12; (a⁵)¹² = a⁶⁰ = 1

Order of a⁶ is 2; (a⁶)² = a¹² = 1

Order of a⁷ is 12

Order of a⁸ is 12/gcd(8,12) = 12/4 = 3

Order of a⁹ is 12/gcd(9,12) = 12/3 = 4

Order of a¹⁰ is 12/gcd(10,12) = 12/2 = 6

Order of a¹¹ is 12

(b) Left coset by 1;

<1,a,a²,....,a¹¹> = <a>

Left coset by a³;

<a³,a⁴,a⁵,a⁶,a⁷, a⁸, a⁹, a¹⁰, a¹¹,1,a,a²> = <a>

Left coset by a⁶;

<a⁶, a⁷, a⁸, a⁹, a¹⁰, a¹¹, 1, a, a², a³,a⁴, a⁵> = <a>

Left coset by a⁹;

<a⁹, a¹⁰, a¹¹,1, a, a², a³, a⁴,a⁵,a⁶, a⁷, a⁸> = <a>

Set of cosets of G by H is <a>

(c) Since G/H is same as G and hence the order of each element is same that we found in part a)