Prove or disprove two groups are nonisomorphic 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.)
 A: 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.
A: 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 $G=GL_2(\mathbb{R})$:

*

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


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


*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.
