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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..

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    $\begingroup$ Stronger (harder) fact: no finite index subgroup of $\mathrm{GL}_{n+1}(\mathbf{Z})$ embeds as a subgroup of $\mathrm{GL}_{n}(\mathbf{Z})$. To prove this one can't make use of finite subgroups, so it needs another approach (an overkill is to use Margulis' results). $\endgroup$
    – YCor
    Jan 6, 2020 at 16:22
  • $\begingroup$ Why do you write $n \geq 2$? I would be really really surprised if $n = 1$ were the exception to the rule and the 2 element group $GL_1(\mathbb{Z})$ were isomorphic to $GL_2(\mathbb{Z})$. $\endgroup$
    – Vincent
    Jan 25, 2020 at 20:47

3 Answers 3

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${\rm GL}(n+1,\mathbb{Z})$ 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).

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    $\begingroup$ You proved a stronger fact: $\mathrm{GL}_{n+1}(\mathbf{Z})$ does not embed as a subgroup of $\mathrm{GL}_{n}(\mathbf{Z})$. $\endgroup$
    – YCor
    Jan 6, 2020 at 16:23
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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.

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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.

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