$GL_{n}(\mathbb{Z})$ 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..

• 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).
– YCor
Jan 6, 2020 at 16:22
• 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 $GL_2(\mathbb{Z})$. Jan 25, 2020 at 20:47

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

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

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.

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.