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?

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    $\begingroup$ I'm not sure what you call "standard left-invariant metric" on $GL_n(\mathbf{R})$. I'm not aware of such a metric (there are plenty of left-invariant Riemannian metric and also some non-Riemannian ones, you have to simultaneously choose one for each $n$) $\endgroup$ – YCor Apr 17 '17 at 15:25
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    $\begingroup$ @YCor the invariant Riemannian metric on $GL(n,\mathbb{R})$ as a Lie group which restrict to $tr(AB^{tr})$ on the tangent space at neutral element. $\endgroup$ – Ali Taghavi Apr 17 '17 at 15:43
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    $\begingroup$ The usual way of generalizing Gromov--Hausdorff convergence to groups is via the Grigorchuk--Gromov space of marked groups. A marked group is a group together with a chosen finite generating set. A marked group corresponds canonically to a normal subgroup of a free group $F_n$. The space of marked groups of rank $n$ is the space of all normal subgroups of $F_n$, with the topology restricted from the natural product topology on the power set of $F_n$. In this topology, the closure of the space of finite groups is the space of groups that are locally emeddable in finite groups (LEF). $\endgroup$ – HJRW Apr 17 '17 at 19:50
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    $\begingroup$ You can certainly define it as a direct limit, though all the papers I know consider fixed $n$. The set of finite groups is always discrete in this topology, since any finitely presented group has a neighbourhood consisting only of (marked) quotients. Here's a reference: arxiv.org/abs/math/0401042 . $\endgroup$ – HJRW Apr 18 '17 at 12:31
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    $\begingroup$ There is a subtlety in the Gromov-Hausdorff definition for spaces I think you are overlooking: the triangle inequality only holds because of the flexibility of metric spaces. Namely, given metric spaces $Y_1,Y_2$ and a compact metric space $X$ that isometrically embeds into both, there is a metric space $Y$ given by gluing $Y_1$ to $Y_2$ along $X$ with the property that $Y_1, Y_2$ isometrically embed. One needs to apply this gluing construction with $X$ the middle space in the triangle inequality in order to verify it. Without such a construction for groups, the triangle inequality need't hold $\endgroup$ – JHance Apr 18 '17 at 15:13

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 $f(x) \leq e^x-1$.

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    $\begingroup$ Thank you very much for your answer. I think you are proving that $d$ is not well defined, since your change an equivalent member. I am surprised:Because I think it is an immediate consequence of the definition that $d$ is well defined.Let $G, G', H $ be three groups with $G \simeq G'$. then $d(G,H)\leq d(G',H)$ and $d(G',H) \leq d(G,H)$. Am I mistaken? $\endgroup$ – Ali Taghavi Apr 18 '17 at 8:44
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    $\begingroup$ @AliTaghavi No, I am proving that it does not satisfy the triangle inequality and hence is not a metric. $\endgroup$ – Will Sawin Apr 18 '17 at 14:27
  • $\begingroup$ May I ask you to more explain on "$p$ different elements of $\mathbb Z/p \mathbb Z$ would have to be within $O(1)$ of the identity"? $\endgroup$ – Ali Taghavi Apr 24 '17 at 14:11
  • $\begingroup$ @AliTaghavi This follows from left invariance of the metric on $GL_n$. We establish that there is a set $S$ in $\mathbb Z/p \mathbb Z \times \mathbb Z/p \mathbb Z$ of size $p$ where every element is within a distance $d=O(1/p)$ of some element of $GL_n(\mathbb R)$, hence every pair of elements is within a distance $2d$ by each other, hence translating by a fixed element $g \in S$, every element in $g^{-1} S \subseteq \mathbb Z/p \mathbb Z \times \mathbb Z/p \mathbb Z$ is in distance $2d$ of the identity. $\endgroup$ – Will Sawin Apr 24 '17 at 14:15
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    $\begingroup$ @AliTaghavi Probably not. It's easy to prove this if you work in $O(n)$ instead of $GL_n(\mathbb R)$, because then the metric is both left and right invariant, but it should still be true in general - one just needs the right inequality. $\endgroup$ – Will Sawin Apr 24 '17 at 18:29

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