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Will Sawin
Will Sawin
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Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Proof. One inclusion is clear so it is enough to prove the other one, by induction on $n$. The case $n\le 1$ is trivial. Suppose $n\ge 2$ and the result proved for smaller $n$. Since $\mathrm{GL}_n(\mathbf{Z})$ is transitive on the set of complete flags, we have $\mathrm{GL}_n(\mathbf{Q})=T_n(\mathbf{Q})\mathrm{GL}_n(\mathbf{Z})$. So it is enough to show that every $A\in T_n(\mathbf{Q})$ normalizing $\Gamma_n(m)$ has the required form.

We see that $Ae_{1n}(m)A^{-1}=e_{1n}(ma_{11}a_{nn}^{-1})$. So $a_{11}/a_{nn}$ is integral and since the same applies to $A^{-1}$ we deduce $a_{nn}=\pm a_{11}$.

The upper left $(n-1)\times (n-1)$ block of $A$ has the same property, so is in $\mathbf{Q}^*\cdot\mathrm{GL}_{n-1}(\mathbf{Z})$. Since its eigenvalues are rational, it follows that all its eigenvalues $a_{11},\dots,a_{n-1,n-1}$ are equal up to sign. Hence, after scalar multiplication, $A$ has only $\pm 1$ on the diagonal, and again by the size $n-1$ case applied to the upper-left and lower-right blocks, all its entries are integral, except maybe the $1,n$ entry. After right multiplication by an upper triangular matrix in $\mathrm{GL}_n(\mathbf{Z})$, we can then suppose $A=e_{1n}(s)$. Then we use that $Ae_{n1}(m)A^{-1}\in\Gamma_n(m)$: its $(1,1)$ entry is $ms$$1+ms$, so $s\in\mathbf{Z}$ and we are done.

Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Proof. One inclusion is clear so it is enough to prove the other one, by induction on $n$. The case $n\le 1$ is trivial. Suppose $n\ge 2$ and the result proved for smaller $n$. Since $\mathrm{GL}_n(\mathbf{Z})$ is transitive on the set of complete flags, we have $\mathrm{GL}_n(\mathbf{Q})=T_n(\mathbf{Q})\mathrm{GL}_n(\mathbf{Z})$. So it is enough to show that every $A\in T_n(\mathbf{Q})$ normalizing $\Gamma_n(m)$ has the required form.

We see that $Ae_{1n}(m)A^{-1}=e_{1n}(ma_{11}a_{nn}^{-1})$. So $a_{11}/a_{nn}$ is integral and since the same applies to $A^{-1}$ we deduce $a_{nn}=\pm a_{11}$.

The upper left $(n-1)\times (n-1)$ block of $A$ has the same property, so is in $\mathbf{Q}^*\cdot\mathrm{GL}_{n-1}(\mathbf{Z})$. Since its eigenvalues are rational, it follows that all its eigenvalues $a_{11},\dots,a_{n-1,n-1}$ are equal up to sign. Hence, after scalar multiplication, $A$ has only $\pm 1$ on the diagonal, and again by the size $n-1$ case applied to the upper-left and lower-right blocks, all its entries are integral, except maybe the $1,n$ entry. After right multiplication by an upper triangular matrix in $\mathrm{GL}_n(\mathbf{Z})$, we can then suppose $A=e_{1n}(s)$. Then we use that $Ae_{n1}(m)A^{-1}\in\Gamma_n(m)$: its $(1,1)$ entry is $ms$, so $s\in\mathbf{Z}$ and we are done.

Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Proof. One inclusion is clear so it is enough to prove the other one, by induction on $n$. The case $n\le 1$ is trivial. Suppose $n\ge 2$ and the result proved for smaller $n$. Since $\mathrm{GL}_n(\mathbf{Z})$ is transitive on the set of complete flags, we have $\mathrm{GL}_n(\mathbf{Q})=T_n(\mathbf{Q})\mathrm{GL}_n(\mathbf{Z})$. So it is enough to show that every $A\in T_n(\mathbf{Q})$ normalizing $\Gamma_n(m)$ has the required form.

We see that $Ae_{1n}(m)A^{-1}=e_{1n}(ma_{11}a_{nn}^{-1})$. So $a_{11}/a_{nn}$ is integral and since the same applies to $A^{-1}$ we deduce $a_{nn}=\pm a_{11}$.

The upper left $(n-1)\times (n-1)$ block of $A$ has the same property, so is in $\mathbf{Q}^*\cdot\mathrm{GL}_{n-1}(\mathbf{Z})$. Since its eigenvalues are rational, it follows that all its eigenvalues $a_{11},\dots,a_{n-1,n-1}$ are equal up to sign. Hence, after scalar multiplication, $A$ has only $\pm 1$ on the diagonal, and again by the size $n-1$ case applied to the upper-left and lower-right blocks, all its entries are integral, except maybe the $1,n$ entry. After right multiplication by an upper triangular matrix in $\mathrm{GL}_n(\mathbf{Z})$, we can then suppose $A=e_{1n}(s)$. Then we use that $Ae_{n1}(m)A^{-1}\in\Gamma_n(m)$: its $(1,1)$ entry is $1+ms$, so $s\in\mathbf{Z}$ and we are done.

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Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.

Proof. One inclusion is clear so it is enough to prove the other one, by induction on $n$. The case $n\le 1$ is trivial. Suppose $n\ge 2$ and the result proved for smaller $n$. Since $\mathrm{GL}_n(\mathbf{Z})$ is transitive on the set of complete flags, we have $\mathrm{GL}_n(\mathbf{Q})=T_n(\mathbf{Q})\mathrm{GL}_n(\mathbf{Z})$. So it is enough to show that every $A\in T_n(\mathbf{Q})$ normalizing $\Gamma_n(m)$ has the required form.

We see that $Ae_{1n}(m)A^{-1}=e_{1n}(ma_{11}a_{nn}^{-1})$. So $a_{11}/a_{nn}$ is integral and since the same applies to $A^{-1}$ we deduce $a_{nn}=\pm a_{11}$.

The upper left $(n-1)\times (n-1)$ block of $A$ has the same property, so is in $\mathbf{Q}^*\cdot\mathrm{GL}_{n-1}(\mathbf{Z})$. Since its eigenvalues are rational, it follows that all its eigenvalues $a_{11},\dots,a_{n-1,n-1}$ are equal up to sign. Hence, after scalar multiplication, $A$ has only $\pm 1$ on the diagonal, and again by the size $n-1$ case applied to the upper-left and lower-right blocks, all its entries are integral, except maybe the $1,n$ entry. After right multiplication by an upper triangular matrix in $\mathrm{GL}_n(\mathbf{Z})$, we can then suppose $A=e_{1n}(s)$. Then we use that $Ae_{n1}(m)A^{-1}\in\Gamma_n(m)$: its $(1,1)$ entry is $ms$, so $s\in\mathbf{Z}$ and we are done.