Proving Cyclic Subgroups of Order n are Isomorphic

Cyclic subgroups play a critical role in group theory, providing a foundation for understanding the structure of groups. In this article, we will explore how to prove that two cyclic subgroups of the same order n are isomorphic.

Introduction to Cyclic Groups

A cyclic group is a group that is generated by a single element. If a group G is generated by a single element x, then it is denoted by x. The order of the cyclic group x is the smallest positive integer n such that x^n e, where e is the identity element of the group.

Proving Isomorphism of Cyclic Subgroups

To prove that two cyclic subgroups X and Y, both of order n, are isomorphic, we need to establish a bijective homomorphism between them. Let's denote X by x and Y by y, where x and y are generators of X and Y respectively. A homomorphism ? can be defined as follows:

?: X → Y ?: xi ? yi

We will now verify that ? is a homomorphism and that it has the properties required for isomorphism.

Homomorphism Property

To show that ? is a homomorphism, we need to prove that:

?(xi middot; xj)  (?(xi)) middot; (?(xj))

Let's substitute the definition of ? and simplify:

?(xij)  yij

We can also write:

(iquest;?(xi)) middot; (?(xj))  yi middot; yj

Since multiplication of elements in a group is associative, we can write:

yij  yi middot; yj

This shows that ? is indeed a homomorphism.

Surjective Property

To prove that ? is surjective, we need to show that for every element yj in Y, there exists an element xk in X such that ?(xk) yj. Since y generates Y, for any j from 0 to n-1, there exists a unique k such that xk yj. Hence, ? is surjective.

Kernel Triviality

To prove that ? is an isomorphism, we need to show that its kernel is trivial. Let's denote the kernel of ? as Ker(?). If xi is in Ker(?), then:

?(xi)  yi  eY

Here, eY is the identity element of Y. Hence, yi eY. Since the order of y is n, we have:

yi  eY rarr; i  kn for some integer k

Substituting yi eY back into xk yj, we get:

xkn  xnk  eX

Since the order of x is n, xn eX. Therefore, kn must be a multiple of n, which implies that k must be an integer multiple of n. However, since xn eX, the only solution is k 0, meaning that xk eX. Hence, the kernel of ? is trivial, meaning ? is injective and thus bijective.

Finite vs Infinite Cyclic Subgroups

The proof works for both finite and infinite cyclic subgroups, but the isomorphism of infinite cyclic subgroups is simpler because the generators of an infinite cyclic group are unique up to multiplication by -1.

Checking for Abelian Property

In addition to isomorphism, it is often useful to check if the subgroups are abelian. For cyclic groups, checking the order of elements can help determine if both subgroups are abelian. If both subgroups contain the same number of elements of the same order, then the subgroups are abelian.

In conclusion, proving that two cyclic subgroups of the same order are isomorphic involves establishing a homomorphism, showing that it is surjective, and proving that its kernel is trivial. This process can be applied to both finite and infinite cyclic subgroups.

Keywords: cyclic subgroups, isomorphic, homomorphism