$GL_{n}(\mathbb{Z})$ and $GL_{n+1}(\mathbb{Z})$ are not isomorphic Can somebody help me on this problem? Thank you!

Show that $GL_{n}(\mathbb{Z})$ and $GL_{n+1}(\mathbb{Z})$ are not isomorphic $\forall n \geq 2$.

My approach:

I tried to create an isomorphism $f : GL_n(\mathbb{Z}) \to GL_{n+1}(\mathbb{Z}), f(X) = Y (X \in GL_n(\mathbb{Z}) \text{ and } Y \in GL_{n+1}(\mathbb{Z})$ but idk how to get a contradiction.. 

 A: ${\rm GL}(n+1,\mathbb{Z})$ has an elementary abelian subgroup $H$ of order $2^{n+1}$ consisting of diagonal matrices with entries $\pm 1$.
Since the irreducible (complex) representations of abelian groups have degree $1$, it is easy to see that $H$ has no faithful complex representation of degree less than $n+1$, so ${\rm GL}(n,{\mathbb Z})$ has no subgroup isomorphic to $H$.
This proves the stronger result that ${\rm GL}(m,{\mathbb Z})$ is isomorphic to a subgroup of  ${\rm GL}(n,{\mathbb Z})$ if and only if $m \le n$ (the if part is easy).
A: Notice that the center of $SL_{\mathrm{odd}}(\mathbb Z)$ is trivial, while the center of $SL_{\mathrm{even}}(\mathbb Z)$ is nontrivial (because $\mathrm{diag}(-1,-1,\dots,-1)\in SL_{\mathrm{even}}(\mathbb Z)$). Thus $SL_n(\mathbb Z)$ is not isomorphic to $SL_{n+1}(\mathbb Z)$. It should not be very hard to derive $GL_n(\mathbb Z)$ is not isomorphic to $GL_{n+1}(\mathbb Z)$ from here.
A: As it is already noticed, $GL(n+1,\mathbb Z)$ has a subgroup isomorphic to $\mathbb Z_2^{n+1}$. If $GL(n,\mathbb Z)\simeq GL(n+1,\mathbb Z)$, then this subgroup corresponds to a subgroup of $GL(n,\mathbb Z)$ consisting of $2^{n+1}$ matrices $A_i$ with $A_i^2=I_n$ and $A_iA_j=A_jA_i$ for all $i,j$. Then the matrices $A_i$ are simultaneously diagonalizable (over $\mathbb C$), that is, there is $S\in GL_n(\mathbb C)$ such that $SA_iS^{-1}$ are diagonal matrices. But $SA_iS^{-1}=\mathrm{diag}\underbrace{\{\pm1,\dots,\pm1\}}_{n\text{ times}}$, and this is impossible since  $SA_iS^{-1}$ are $2^{n+1}$ distinct matrices.
