The completion of the space of finite groups Edit: I revise the  question based on the comment  conversations
Let $\mathcal{F}$ be  the set of  all equivalence classes  of  finite  groups under the "Isomorphism" equivalence relation.
We  define a  pseudo metric  $d$  on $\mathcal{F}$ as  follows:
$$d(G,H)= \inf \{Hd(\tilde {G}_{n},\tilde{H}_{n})\} $$
where  $\inf$ is taken  over all arbitrary   isomorphic  copies $\tilde{G}_{n}$  and  $\tilde{H}_{n}$ of  $G$ and  $H$ in   $Gl(n,\mathbb{R})$, respectively, while $Hd$ is the Hausdorff distance   in $GL(n,\mathbb{R})$ induced by its standard left  invariant  metric.
The  definition of  this  metric  is  motivated  by the  Hausdorff Gromov  metric  on the  space  of  compact  Riemannian  manifolds.

Is $d$   a  metric  on $\mathcal{F}$? If  the  answer  is  yes, we  denote by  $\bar{\mathcal{F}}$ the   completion of $\mathcal{F}$. What  can be  said about  an object $Z$ in  $\bar{\mathcal{F}}$? 

Can one  consider  the unit circle, in some reasonable  sense, as  an  object in this completion?
Is there  a  natural group  structure  on every element $Z\in \bar{\mathcal{F}}$? Is there  a  natural topology on $Z$?
Is  $\bar{\mathcal{F}}$ a  compact space?
 A: i don't think it's a metric. Take a large prime $p$. By embedding $\mathbb Z/p\mathbb Z$ and $\mathbb Z/(p^2+p)\mathbb Z$ in the circle $S^1 \subseteq GL(2,\mathbb R)$, one sees that the distance between them is at most $O(1/p)$. By embedding $\mathbb Z/p \mathbb Z \times \mathbb Z/p \mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ in the torus $S^1 \times S^1 \subseteq GL(4,\mathbb R)$, one again sees that the distance between them is at most $O(1/p)$.
But of course $\mathbb Z/(p^2+p)$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ are isomorphic, so one would be forced to conclude by the triangle inequality that the distance between $\mathbb Z/p\mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ is $O(1/p)$.
But this is false. If they had that distance in some $GL(n,\mathbb R)$, then by pidgeonhole, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1/p)$ of some element of $\mathbb Z/p\mathbb Z$ and hence within $O(1/p)$ of each other. By left invariance, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1)$ of the identity.
But in any representation of $\mathbb Z/p\mathbb Z \times \mathbb Z/p\mathbb Z$, only $o(p)$ elements have eigenvalues within $o(1/\sqrt{p})$ of the identity, since we can write the representation as a sum of characters, the eigenvalues on each character must be $p$th roots of unity, and each element is determined by its eigenvalues on two independent characters.
So we just need to check that every element within $O(1/p)$ of the identity has eigenvalues within $o(1/\sqrt{p})$ of the identity.
In fact   we  can show more is true, and  an element within $d$ of the identity matrix can't move any vector of length one by a distance of greater than $e^{d}-1$. Since $e^{O(1/p)}-1 = O(1/p) = o(1/\sqrt{p})$, we obtain the desired conclusion. To check this, differentiate $Mv$ with respect to $M$ and observe that its operator norm with respect to your metric is the operator norm of $M$, so if $f(x)$ is the maximum total distance moved a vector of length one by a matrix within $x$ of the identity, $df/dx \leq 1+f$ so $f(x) \leq e^x-1$.
