Are these groups of order $81$ isomorphic?

The classification of groups of order $p^4$ is well known. However, different sources sometimes classify in different way. I am trying to compare, in the case $p=3$ between Burnside list (p. 100-102) http://www.gutenberg.org/files/40395/40395-pdf.pdf?session_id=730cced368b475738bfcf2c60e81ebba6fd01e04

and the smallgroups http://users.minet.uni-jena.de/~green/Coho/81gps/index.html

Now, I am trying to find a group in Burnside list which isomorphic to the group $4$ in the smallgroups. I think that it can only be the group $(viii)$, however I hope someone can check and tell me if I am right. Thanks in advance.

The information in the second link you quote (until the first rule) does not appear to be enough to reconstruct the isomorphism class of the group, which I will call $G_{s}$, as it might well be abelian.
Assuming $G_{s}$ is not abelian, then the centre must have order $3^{2}$. (Any larger, and the group will be forced to be abelian.) So the group has nilpotence class $2$. If $a, b$ are generators, then, the derived subgroup will be $\langle [a, b] \rangle \le Z(G_{s})$, of order $3$. (Here $d = [a, b] = a^{-1} b^{-1} a b$.) By regularity, if you wish, $d$ is a third power, say $d = a^{3}$ (renaming $a, b$ in case). Now you see that this forces $G_{s}$ to be isomorphic to the group $G_{b}$ you name in Burnside's table, as both $G_{s}$ and $G_{b}$ are described by the relations $$a^{9}, b^{9}, b^{-1} a b = a^{4}.$$
• Thanks a lot. Indeed, the group is not abelian. I think it is clear since there are $4$ maximal subgroups but maybe I am wrong. – Ofir Schnabel Apr 29 '15 at 8:45
• @OfirSchnabel, you're welcome, and thanks. Group number 2 in the small groups list is abelian, and has also $4$ maximal subgroups. But it is clear from the context (and no doubt from the further information provided in the page) that the group is non-abelian. – Andreas Caranti Apr 29 '15 at 8:53
• You are right again about group number $2$. – Ofir Schnabel Apr 29 '15 at 8:58