TITLE: What does it mean for something to hold "up to isomorphism"?
QUESTION [39 upvotes]: For example, to say that there are 2 such groups up to isomorphism such that the order of G is equal to $p^2$?

REPLY [5 votes]: A quick thought I'd like to add to the other answers that one needs to keep in mind.
Sometimes we get so caught up in the equivalence defined by isomorphism that we actually forget that two isomorphic groups can be different groups and sometimes this makes quite a difference to one's discussion. For example, a group can have two subgroups which are isomorphic copies of one another within the supergroup, but this is quite different from the groups being the same. In relation to the supergroup, two isomorphic copies of a subgroup can have quite different properties: consider $\mathbb{R}\times\mathbb{T}^2$, the product of the real line and the torus, where group entities can be represented as 3-tuples $(x,\,e^{i\,\phi},\,e^{i\,\theta})$ and the group operation is defined by:
$$(x_1,\,e^{i\,\phi_1},\,e^{i\,\theta_1})\,(x_2,\,e^{i\,\phi_2},\,e^{i\,\theta_2}) = (x_1+x_2,\,e^{i\,(\phi_1+\phi_2)},\,e^{i\,(\theta_1+\theta_2)})$$
The the subgroups:
$$G = \{(x,\,1,\,1)|\,x\in\mathbb{R}\}$$
$$H(\alpha) = \{(0,\,e^{i\,\phi},\,e^{i\,\alpha\,\phi})|\,x\in\mathbb{R}\}$$
where $\alpha$ is irrational are isomorphic copies of one another: they are both isomorphic to the real additive group $(\mathbb{R},\,+)$. But $G$ is closed in the obvious topology of $\mathbb{R}\times\mathbb{T}^2$ (i.e. that for which a base is all sets of the form $(\mathcal{I}_1,\,e^{i\,\mathcal{I}_2},\,e^{i\,\mathcal{I}_3})$ where the $\mathcal{I}_j$ are open intervals in $\mathbb{R}$), whereas $H(\alpha)$ is not: its closure is the whole torus $\mathbb{T}^2$.

(to prove that $H(\alpha)$ is isomorphic to $(\mathbb{R},\,+)$, assume that there is a member $(0,\,e^{i\,\phi},\,e^{i\,\alpha\,\phi})$ equal to the identity $(0,1,1)$ other than the identity itself, i.e. that the "thread" $H(\alpha)$ links up with itself: then $\phi=2\,m\,\alpha\,\phi = 2\,n\,\pi$ for $m,\,n\in\mathbb{Z}$, contradicting the assumption that $\alpha$ was irrational)