Show that the number of groups of order 8181 is equal to the number of groups of order 81 (up to isomorphism) 
Show that the number of nonisomorphic groups of order $8181=3^4\cdot 101$ is equal to the number of nonisomorphic groups of order $81$. Find all the abelian groups of order $8181$ and at least one that is not abelian.

the second part (finding all abelian groups) doesn't seem very hard. But I don't know how to approach the first part. I tried applying the Sylow theorems but it didn't seem to help.
 A: For the first part we proof that each such group $G$ is isomorphic to a direct product of the cyclic group $\mathbb Z_{101}$ of order $101$ and a group $H$ of order $81$, i.e.,
$$ G \cong H \times \mathbb Z_{101} $$
By Sylow's first theorem there exists a $101$-Sylow subgroup $G_{101}$ of order $101$. Let $n_{101}$ denote the number of $101$-Sylow subgroups then by Sylow's third theorem $n_{101} | 81$ and $n_{101} \equiv 1 \; (\mathrm{mod}\; 101)$. It follows that $n_{101} \leq 81$ and therefore $n_{101} = 1$. Since all conjugates of a $p$-Sylow are $p$-Sylow themselves, $\mathbb Z_{101} \cong G_{101}$ is normal in $G$.
Now let $G_{81}$ be a $3$-Sylow subgroup of order $81$ in $G$. Again this exists by Sylow's first theorem. As before we denote by $n_{3}$ the number of its conjugates of $G_{81}$ and therefore the number of $3$-Sylow subgroups in $G$. We have $n_{81} | 101$ and $n_{81} \equiv 1 \; (\mathrm{mod}\; 81)$. Again this leaves no other choice but $n_{81} = 1$ and $G_{81}$ is normal in $G$.
Finally, let $a \in G_{81} \cap G_{101}$ then the order of $a$ must divide $81$ and $101$ and since these are relative prime, the intersection is trivial. We conclude, that
$$ G = G_{81} \times G_{101} \cong H \times \mathbb Z_{101},$$
where $H$ denotes a group of order $81$. This implies that there at least as many groups of order $81$ as there are groups of order $8181$. On the other hand, using this direct product we can easily construct a group of order $8181$ from a given group $H$ of order $81$. So the numbers must coincide.
Finding the abelian groups of order $8181$ is a direct application the fundamental theorem of finitely generated abelian groups.
As for the example of a non-abelian group pick any non-abelian group of orded 81, e.g., any of these. And define $G$ as its direct product with $\mathbb Z_{101}$.
A: Both the Sylow $3$-subgroup and the Sylow $101$-subgroup are normal, so a group $G$ of order $8181$ is a direct product $A\times B$ where $A$ is of order $81$ and $B$ is of order $101$. There is a unique isomorphism class of groups of order $101$, so a group of order $8181$ is just the direct product of a group of order $81$ with a cyclic group of order $101$. The claim follows.
To find a nonabelian group, it's enough to find a nonabelian group of order $27$. There is one of those, it's a standard example of a finite group with exponent $3$ that is not abelian.
