# 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?

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.
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?