Proving that $\text{SL}_2 (\mathbb{Z})$ is closed under inverses

Question

I am trying to verify that $\text{SL}_2 (\mathbb{Z})$ is a group under matrix multiplication. The only property I am not certain on is closure under inverses.

Given $A \in \text{SL}_2 (\mathbb{Z})$, I know that $A^{-1}$ exist (it has non-zero determinant). I need to show that $A^{-1}$ has determinant $1$ and integer entires. It's true in general that $\det(A^{-1}) = \frac{1}{\det(A)}$, so if $\det(A) = 1$, then $\det(A^{-1}) = 1$.

Proving that $A^{-1}$ has integer entries is a bit tougher, because I don't think it is necessarily true that if $\det(A^{-1}) = 1$, then all of its entries are integers (though I struggle to think of a counterexample). The only proof I could think of involves the adjugate matrix, $\text{adj}(A)$. In general, we have $$ A \text{adj}(A) = \det(A) I$. $$ As $\det(A) = 1$, the inverse of $A$ is the adjugate matrix. The adjugate matrix is the transpose of the matrix of cofactors, whose entries are partial determinants around entry $(i,j)$, up to a sign. The entries of $A$ are integers, so every entry of the matrix of cofactors is an integer ($\mathbb{Z}$ is closed under addition and multiplication), so the adjugate matrix contains only integers.

Is this reasoning sound?

Answer 1

The argument with the adjugate matrix is correct and using it will get you to the heart of the matter: Any $n\times n$ matrix $A$ with entries in any commutative ring $R$ is invertible (in $R^{n\times n}$) if and only if $\det(A)$ is a unit in $R$.

Having established this, it is then clear that $\mathrm{SL}_n(R)$ as the kernel of $\det$ only contains invertible matrices and is therefore a group by your previous arguments.

Answer 2

You can just use the (unique) explicit form of the inverse of a $2 \times 2$ matrix, which is given by $$ A^{-1}=\frac{1}{\det(A)} \begin{pmatrix}d & -b \\ -c & a \end{pmatrix} $$ and this has integer entries if $A$ has only integer entries and $\det(A)= \pm 1$.

Answer 3

Here's another approach for proving that $SL_n(\mathbb Z)$ is closed under inverses. Let $A\in SL_n(\mathbb Z)$. Consider the characteristic polynomial of $A$, $p_A(x)=\text{det}(xI_n-A)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$. Note that $a_0=p_A(0)=det(-A)=(-1)^ndet(A)=(-1)^n$ and $a_i\in\mathbb Z$. By Cayley-Hamilton theorem it follows that $p_A(A)=0$, and hence $$0=A(A^{n-1}+a_{n-1}A^{n-2}+\ldots+a_1I_n)+(-1)^nI_n,$$ which implies that $$A^{-1}=(-1)^{n+1}(A^{n-1}+a_{n-1}A^{n-2}+\ldots+a_1I_n).$$ Then necessarily $A^{-1}$ has integer coefficients.

Answer 4

Proving that $A^{−1}$ has integer entries is a bit tougher, because I don't think it is necessarily true that if $\det(A^{-1})=1$, then all of its entries are integers (though I struggle to think of a counterexample).

This logic is not valid. There is no such counterexample. If $A$ is in $SL_2(\Bbb Z)$, then we cannot find a counterexample, because actually $SL_2(\Bbb Z)$ is a group, as we know.

If $A\in M_2(\Bbb Z)$ just satisfies $\det(A^{-1})=1$, then we have $\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$, so that $\det(A^{-1})=1$ implies also $\det(A)=1$, i.e., $A\in SL_2(\Bbb Z)$. So there is no counterexample.

If you drop the condition that $A$ has integer coefficients, and only require $\det(A^{-1})=1$, then of course you can take $$ A=\begin{pmatrix} \frac{1}{2} & 0 \cr 0 & 2\end{pmatrix}. $$