Say we wish to show that a given group $G$ is described by the presentation $ H = \langle g_1,g_2,\cdots \mid r_1 = r_2 = \cdots = 1\rangle$. My line of reasoning is as follows:
(1) Show by our knowledge of $G$ that suitable generators and relations exist, implying that $|H| \geq |G|$.
(2) Show by some other means that $|H| \leq |G|$. (For example given the relations on $H$ we show that all words in $H$ are necessarily equal to one of a set of $n = |G|$ words).
(3) We now know that $|H| = |G|$.
Can we now conclude that the groups are actually isomorphic, knowing that their orders are the same? Or is some additional information needed?
In general, knowing that the orders are the same is very poor information if you've got nothing else.
However, the general way to show that a presentation is a presentation is by defining an isomorphism, which means three things:
Now if you know that the orders are the same, and if the order is finite, then injectivity is enough, and surjectivity is enough.
However, proving both injectivity and surjectivity might turn out to be easier than showing that the orders are the same to begin with.
If the order is infinite, then knowing that it's the same for both groups is essentially useless.