Consider $A=\langle a\rangle$, cyclic group of order $9$ and $B=\langle b\rangle$, cyclic group of order $3$.
Consider now the following action of $B$ on $A$ via automorphism: $$\gamma:B\longrightarrow Aut(A)$$ defined by $b\mapsto\gamma_b\in Aut(A)$ where $\gamma_b(a):=a^4$ (and this completely defines $\gamma$); moreover there is a theorem that says that every automorphism is essentially a conjugation (see Martin Isaac, Finite Group Theory, p.69) hence we can denote $\gamma_b(a)$ with $a^b$. So the action is totally defined by $a^b=a^4$.
Hence we can consider the semidirect product $G=A\rtimes_{\gamma}B$; this is a group of order $|A||B|=27$, it contains an element of order $9$, that is $(a,1)$ and an element of order $3$ that is $(1,b)$ (as a set, $G=A\rtimes_{\gamma}B$ is equal to $A\times B$, and the operation we put on $A\rtimes_{\gamma}B$ to make it a group is $(a_1,b_1)*_{\gamma}(a_2,b_2):=(a_1\gamma_{b_1^{-1}}(a_2),b_1b_2)$; of course the notion of order is with respect this operation!); moreover these two elements respect the following relation: $(a,1)^{(1,b)}=(a,1)^4$ (lhs in the sense of conjugation, rhs in the sense of power!).
Consider now $G':=\langle a,b\;:\;a^9=b^3=1, a^b=a^4\rangle$. This group respects the same properties as the previous one (it has an element of order $3$ that is $b$, one of order $9$ that is $a$ and they satisfy $a^b=a^4$). The unique thing I can't say is that $|G'|=27$ (how can I compute it a priori?). Apart from cardinality, I have two groups that are totally characterized by the same two elements which are in the same relation one with the other, and no other request is needed in the definition of both. Is this sufficient to say that $G\simeq G'$? Is there a way to tell it in a smarter way?
Thank you all!