In order to show that any two cyclic groups of finite order are isomorphic to each other, it suffices to show that every cyclic group of finite order is isomorphic to the group of integers modulo with the operation of addition modulo , where .

Towards this end, it can be shown that the function defined as is a group isomorphism. Recalling the definition of group isomorphism, it suffices to show that is a bijective homomorphism. It is straightforward to check that is a bijective function. The fact that is a homomorphism is proven also easily using laws of exponents for groups, as follows: .

Proving the homomorphic property of as above works well if is taken to be that is from the additive group of integers to the cyclic group of infinite order generated by . However, in the case of with , there is a slight oversight; the addition implied by is addition modulo .

Let’s denote the addition modulo by . So, now the proof of homomorphic property takes the form . The identity is not the same as; the former is one of the common exponent laws for groups, while the latter is not.

What needs to be shown is that the exponents of addition and of addition modulo coincide, i.e. . The proof of this identity is simple, yet it is useful to be aware of such potential oversight.

According to the division algorithm for integers, there exist unique integers and such that and . Thus, . Since the cyclic group is of order , it holds that , where is the identity element of . So, .

Notice that coincides with the result of the addition modulo , that is . It follows that , which completes the proof.

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