When are same order groups are isomorphic So the question is as follow:

$G,H$ are two groups of order $n=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ are G,H isomorphic if:
(a) $G,H$ have to same composition factors
(b) $G,H$ are solvable
(c) $G,H$ are nilpotents
(d) $G,H$ are abelian
(e) $G,H$ are cyclic

So I've reached the follow conclusions:
for (d) the answer is now because we can take $\mathbb Z/n\mathbb Z$ and $\mathbb  Z/2\mathbb Z \times \mathbb  Z/3\mathbb Z \times\mathbb  Z/5\mathbb Z \times\mathbb  Z/7\mathbb Z \times\mathbb  Z/11\mathbb Z \times\mathbb  Z/13\mathbb Z $  which aren't isomorphic.
for (e) we can define a function from one group's generator to the other and it'll be isomorphism, so yes.
Now I'm having trouble with (a) - (d) though I saw a theorem I'm not sure how to prove that states the if $G,H$ are solvable with the same order then they have the same composition factors which makes (b) dependent on (a). any help with the rest will be appreciated thanks!

false up to b, true from c.
For c note that a nilpotent group is the direct product of it's sylow subgroups. For a counterexample to b see the comments in the post given by @HSN - the counterexample is $S_3\times C_5\times C_7\times C_{11}\times C_{13}$.
You're mistake on d is that the second group you mention is cyclic (generated by $(1,1,\ldots,1)$).
