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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?
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up vote 4 down vote accepted
  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 $\mathbb{Z}_{p^3}$.
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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
@Flast9 There's no absolute way which works for every presentation - it's like solving a puzzle. Sometimes group presentations can be very complicated. In general, I try to figure out the commutators between each generator, and see if you can deduce the order of every element. – Alexander Gruber Oct 29 '12 at 22:56

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