Does knowing the generators and the orders of those generators of a finite group $G$ completely determine $G$? [closed]

Does knowing the generators and the orders of those generators of a finite group $G$ completely determine $G$?

For instance, is there only one group $G$ generated by the elements $\{g,h\}$ where $|g|,|h| = 3$?

• Perhaps up to isomorphism. – IAmNoOne Dec 20 '16 at 23:14
• For the additional information needed to uniquely define a finite group in terms of its generators, you might be interested in the notion of a group presentation. – hardmath Dec 21 '16 at 1:21
• @Chocolate Sorry but the Klein 4-gruppe is $C_2 \times C_2$ see this wikipedia article. – Marc Bogaerts Dec 21 '16 at 17:48
• @BogaertsMarc whoops well that was a dumb comment on my part! Thanks >_< – ChocolateAndCheese Dec 21 '16 at 19:40

There can be more than one group with the same number of generators and with the generators being of the same order. For example $Z_4\times Z_2$ and $D_8$ both have two generators, one of order $2$ and one of order $4$.
Consider the Quaternions: this group can be generated by an element of order $2$ and three elements of order $4$. The same can be said of $\mathbb{Z}_2\oplus\mathbb{Z}_4\oplus\mathbb{Z}_4\oplus\mathbb{Z}_4$. The former is not Abelian, the latter is.
EDIT: In the example you gave, $\mathbb{Z}_3\oplus\mathbb{Z}_3$ is of course one such group. There is another, albeit trivial case of $\mathbb{Z}_3$ where $g=1,h=2$. Both elements have order $3$, but these two groups are distinct. You could argue that this is cheating, since $\mathbb{Z}_3$ only needs one of those two generators, but you can still express it as $\mathbb{Z}_3=<g,h| g^3=h^3=gh=e>$