Consider the factor groups $(Z_2 \times Z_4)/{<}(0,2){>}$ and $(Z_2 \times Z_4)\times {<}(1,2){>}$. I wish to classify the two groups according to the fundamental theorem of finitely generated abelian groups, thus finding an isomorphism. Consider the first group. We have that ${<}(0,2){>} = \{(0, 2), (0,0)\}$. Therefore the factor group is of order $4$. Since the first remains unchanged, and there is but two elements in the second, $Z_2$ remains unchanged while $Z_4$ "collapses" to $Z_2$, and we are left with $Z_2 \times Z_2$ as an isomorphism.
Moving on to the second factor group, we have that ${<}(1,2){>} = \{(1,2),(0,0)\}$, thus this group is also of order $4$. Seeing as both elements change, we may not have that the group is isomorphic to $Z_2 \times Z_2$, which leaves us with $Z_4$ as the only viable option.
Note that my arguments above are the result of me stubbornly attempting to be rigorous: my original argument was that I guessed the first isomorphism, and since the second group was of the same order, I assumed that there was some trick and the answers were different. I do not fully understand the concepts used here. What does it mean for something to "collapse"? How can one use these arguments to prove that there is so isomorphism from the second group to $Z_4?$
EDIT: Solving another one of this task, I have gained, if not understanding, some kind of rhythm. Consider the factor group $(Z_4 \times Z_8)/{<}(1,2){>}$. We have that this is a group of order 8, and since the $1$ exhausts all of $Z_4$, we have that $Z_4$ collapses and we are left with $Z_8$ as an isomorphism. Is this correct?
Best Answer
I take it you want to show there is an isomorphism from the second group to the cyclic group of order $4$. So you want to find an element of order $4$ in the quotient group. That will have to come from an element of order $4$ in the big group. One such element is $(0,1)$. Now just check to see that that element indeed has order $4$ in the quotient group.