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?
Best Answer
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.