# Give a reason why $\mathbb Z_{64}\times \mathbb Z_4$ and $\mathbb Z_{64}\times \mathbb Z_{2}\times \mathbb Z_{2}$ are not isomorphic.

The question is about classifying groups of order $$256$$ with at least one element of order $$64$$, and justify why the elements of the list are non-isomorphic. I'm done except for showing that $$\mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_{2}\text{ and } \mathbb Z_{64}\times \mathbb Z_{4}$$ are not isomorphic. Of course they are not, that's the content of the classification theorem, but the question begs a direct approach.

All similar questions I found on MSE differentiate groups of this kind from order elements, e.g. $$\mathbb Z_2\times \mathbb Z_2$$ and $$\mathbb Z_4$$ are not isomorphic as only the latter contains an element of order 4. We can't apply that here.

My solution: the order of a an element of a product is the minimum common multiple of the non-zero orders. Thus an element of $$\mathbb Z_{64}\times \mathbb Z_{2}\times\mathbb Z_2$$ has order 2 iff it's nonzero and its $$\mathbb Z_{64}$$ factor is $$0$$ or $$32$$ (the elements of order 2 in $$\mathbb Z_{64}$$). We get a total of 7 elements of order 2, $$(32,0,0),(0,1,0) ,(0,0,1) ,(0,1,1) ,(32,1,0),(32,0,1) \text{ and }(32,1,1).$$ On the other hand, the same reasoning shows that there are only 3 elements of order 2 in $$\mathbb Z_{64}\times\mathbb Z_{4}$$, namely $$(32,0), (0,2) \text{ and }(32,2).$$

Is this correct? I think so, but I spent more time in this question than I'd like to admit, that's why I'm posting it. Any other approaches?

• It is correct and you don't need anything else. Good job. Yet Mark Bennet has a very nice idea in his answer: one of the groups has $\;\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2\;$ as a subgroup, whereas the other one has not ... Commented Dec 31, 2020 at 22:24

Depending on where you are in your learning you might be expected to show that if the first component is other than $$0$$ or $$32$$ the order is greater than $$2$$ (and similarly for the second component in the $$\mathbb Z_4$$ case). But the reasoning is completely sound.
You might also identify a subgroup $$\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$$ in one case rather than the other. Orders of elements and subgroups are two potentially distinguishing features to look out for.
Draw the lattice of subgroups, maybe. The lattice of the Klein four group is not a total order, but that of the cyclic group of order four is. Then note how doing $$G\times -$$ changes the lattice.