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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 $|g|,|h| = 3$?

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  • $\begingroup$ Perhaps up to isomorphism. $\endgroup$ – IAmNoOne Dec 20 '16 at 23:14
  • $\begingroup$ 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. $\endgroup$ – hardmath Dec 21 '16 at 1:21
  • $\begingroup$ @Chocolate Sorry but the Klein 4-gruppe is $C_2 \times C_2$ see this wikipedia article. $\endgroup$ – Marc Bogaerts Dec 21 '16 at 17:48
  • $\begingroup$ @BogaertsMarc whoops well that was a dumb comment on my part! Thanks >_< $\endgroup$ – ChocolateAndCheese Dec 21 '16 at 19:40
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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 $4$.

Occasionally this information will point to a single group, but not always.

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  • $\begingroup$ Thanks. What extra information would be needed to completely determine the group then? $\endgroup$ – Joshua Lin Dec 20 '16 at 23:23
  • $\begingroup$ @JoshuaLin there are a lot of ways you can completely determine a finite group. The most straightforward is to give the multiplication table for the entire group. To see more concise methods, this might help: en.wikipedia.org/wiki/Presentation_of_a_group $\endgroup$ – Sean English Dec 20 '16 at 23:29
  • $\begingroup$ @JoshuaLin: An idea of the lack of uniqueness, even when the order of the commutator of the two generators is known, can be gathered from the paper "Groups Defined by the Order of Two Generators and the Order of their Commutator" by G. A. Miller (1908). $\endgroup$ – hardmath Dec 21 '16 at 1:16
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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 $\mathbb{Z}_3=<g,h| g^3=h^3=gh=e>$

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