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.

Question

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?

Answer 1

That was my first thought of a way to do it.

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.

Answer 2

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.