Question

Suppose that G is abelian group of order 16, and in computing the orders of its elements, you come across an element of order 8 and 2 elements of order 2. Explain why no further computations are needed to determine the isomorphism class of G. provide explanation please.

Answer 1

Here G is an abelian group of order 16.

For any abelian group G of order p^{​​​​​​n} ​​​​​, we have

for some natural numbers summing to n.

We are given G of order 16, that is 2⁴ .

Here p=2 and n= 4.

In computing the orders of its elements, when we come across an element of order 8 and 2 elements of order 2.

Hence we have a cyclic group of order 8 or 2³ ​ generated by the element of order 8, which is isomorphic to Z_{2^{​​​​​3}} and a cyclic group of order 2, which is cyclic to Z_{​​​​​​2} .

As we know 3+1 = 4.

Therefore

and we dont need further computations to determine the isomorphism class of G.