Question

Justify that a group of others 15 is cyclic​

Answer 1

Answer:

In group theory, a branch of abstract algebra, a cyclic group or monogenous group is a group that is generated by a single element.[1] That is, it is a set of invertible elements with a single associative binary operation, and it contains an element g such that every other element of the group may be obtained by repeatedly applying the group operation to g or its inverse. Each element can be written as a power of g in multiplicative notation, or as a multiple of g in additive notation. This element g is called a generator of the group.[1]

Every infinite cyclic group is isomorphic to the additive group of Z, the integers. Every finite cyclic group of order n is isomorphic to the additive group of Z/nZ, the integers modulo n. Every cyclic group is an abelian group (meaning that its group operation is commutative), and every finitely generated abelian group is a direct product of cyclic groups.

Every cyclic group of prime order is a simple group which cannot be broken down into smaller groups. In the classification of finite simple groups, one of the three infinite classes consists of the cyclic groups of prime order. The cyclic groups of prime order are thus among the building blocks from which all groups can be built.

Answer 2

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Together with the identity element which has order 1, that makes 1+12a+10b=15 for some positive integers a and b. This is impossible. Hence by Proof by Contradiction it follows that G must be abelian. Since 15 is a product of 2 distinct primes, by Abelian Group of Semiprime Order is Cyclic, G is cyclic.