Let G be a multiplicative group of $n\times n$ matrices. If $Int(G)\not = \emptyset$ then $G$ is open on $\mathbb{R}^{n^2}$ 
Let G be a multiplicative group of $n \times n$ matrices. If $\text{int}(G)\not = \emptyset$ then $G$ is open on $\mathbb{R}^{n^2}$

Clarifying notation: $\text{int}$ denotes the set of interior points of $G$. If $G$ is an open set then $G = \text{int}(G)$.
I think the best way is to try a contradiction. Suppose that exists $a \in G$ s.t. $a \not \in \text{int}(G) \implies \forall \; \delta>0, B(a,\delta)\cap(\mathbb{R}^{n^2}-G)\not =\emptyset$ which means any ball centered in $a$ must contain points of the complement of G.
Now, if $\text{int}(G) \not =\emptyset\implies \exists b \in G$ s.t. $B{(b,r)}\subset G$ for some $r \in \mathbb{R}_+$. Since $G$ is a multiplicative group, there is an inverse of every $y\in B(b,r)\subset G$.
then $\forall y \in B(b,r)$, $y\times a \in G$ (by the closure axiom on the group axioms).
Now, that's where I'm stuck, my intention was to somehow use the inverse of every $y$ to show that the ball around a must be a subset of $G$ hence getting that $a \in \text{int}(G)$, but I don't know how. Any tips?
 A: You're pretty close: you want to "move" the point $y$ to be near $a$, but you can't just multiply it by $a$; you have to multiply it by the inverse of $b$, too. However, it's easier not to wrap it in a proof by contradiction.
Suppose that $B(b,r) \subseteq G$. Note that "multiplication by $b$" is a homeomorphism (with inverse "multiplication by $b^{-1}$). It follows that $b^{-1}B(b,r)$ is also open, and it is an open set containing $I_n$. And then for any $a \in G$, also $ab^{-1}B(b,r)$ is an open set, now containing $a$. Since every point of $G$ has a neighbourhood in $G$, the set $G$ is open.
A: You're very close. Here's how I would proceed: let $a\in G$ and $b \in \textrm{int}(G)$ with its associated ball $B(b,r)$. Since the multiplication action of $G$ on itself is transitive, there is a $g\in G$ such that $gb = a$. Now, since translation is a homeomorphism in a topological group, $gB(b,r)$ is an open set containing $a$. Show $gB(b,r)$ lies in $G$ and you're done!
I should note that the solution has nothing to do with the fact that $G$ is a group of matrices. Can you generalize?
