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

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 arelocally emeddable in finite groups(LEF). $\endgroup$ – HJRW Apr 17 '17 at 19:50