Prove or disprove two groups are nonisomorphic

Question

This might be a really silly question, but how can one prove (or disprove) that $GL_2(\mathbb{R})$ and $GL_3(\mathbb{R})$ are nonisomorphic as groups? (Of course we know that they are nonisomorphic if we impose more structure. For example, if we view $GL_2(\mathbb{R})$ and $GL_3(\mathbb{R})$ as Lie groups, it's easy to see via a dimension argument that they are not isomorphic as Lie groups.)

Answer 1

For these specific groups, one can use representation theory of finite groups. That is, we can show that $GL_3(\mathbb{R})$ has different finite subgroups to $GL_2(\mathbb{R})$. If we can find a finite group that has a faithful $3$ dimensional real representation, but no faithful $2$ dimensional real representation, then we are done, so it suffices to find a finite group like this, and one can check using character theory that $A_4$ the alternating group on $4$ elements has this property.

Answer 2

This is another way to solve this problem.

An easy computation shows that there are exactly three possibilities for centralizers of elements of the group $G=GL_2(\mathbb{R})$:

1. If $a=\alpha I$, then $C_G(a)=GL_2(\mathbb{R})$;

2. If $a$ is diagonalizable over the complex numbers, then $C_G(a)$ is an abelian group;

3. If $a$ is of the form $$ a=\left( \begin{array}{cc} \alpha & 1 \\ 0 & \alpha \\ \end{array} \right), $$ then $C_G(a)$ is the group of upper triangular matrices.

In the first case the second commutant of $C_G(a)$ is $SL_2(\mathbb{R})$ and is therefore non-Abelian. In the second and third cases the second commutant of $C_G(a)$ is trivial.

Now consider the matrix $a\in GL_3(\mathbb{R})$ $$ a= \left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right). $$ We have $$ C_{GL_3(\mathbb{R})}(a)= \left( \begin{array}{ccc} x & y & z \\ 0 & u & v \\ 0 & 0 & x \\ \end{array} \right),\ x,y,z,u,v\in\mathbb{R},\ x,u\neq0. $$ The second commutant of this group is an Abelian group consisting of matrices of the form $$ \left( \begin{array}{ccc} 1 & 0 & z \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right),\ z\in\mathbb{R}. $$ It follows that the groups $GL_2(\mathbb{R})$ and $GL_3(\mathbb{R})$ are non-isomorphic.