If $a^{3^3}=b^9=1$ for generators $a,b$ of $G$, can we conclude that $G$ is a $3$-group?

I know that $a$ and $b$ are generators of a group $G$ and $a^{3^3}=b^9=1$.

1. Are these informations sufficient to affirm that the group is a $3$-group?
2. Adding the relation $b^{-1}ab=a^4$, can we state that it is a $3$-group?
-

1. No. We can just keep taking elements $ab$, $abab$, $ababab$, etc. and there is nothing in the relations to stop us. Without relating $a$ and $b$ somewhere, the group is infinite.
2. Now, there is enough to tell that it is a $3$-group - in fact, it is the Sylow $3$-subgroup of the holomorph of $\mathbb{Z}_{27}$. This is a special case of the following characterization: whenever you have $\langle a,b | a^{p^3}=b^{p^2}=1, a^b=a^{p+1}\rangle$, this presentation describes the Sylow $p$-subgroup of the holomorph of $\mathbb{Z}_{p^3}$.
Thank you, but there is a question I like to ask. You already had the knowledge of a $3$-group (the one you mentioned) and say that that group satisfies the requirements, ok. But suppose you were not knowing the group you talk about: is there a method that allows us to say that the relations above (including the one in point 2) define a $3$-group? – Flast9 Oct 29 '12 at 19:36