Can any finite group of order $n$ be embedded in $SL_n(\mathbb{Z})$?

I have found the answer to my question so I thought I should share.

The main idea is that once we can embedd $S_n$ in $GL_{n-1}(\mathbb{Z})$, it is easy to create a injective morphism from $GL_{n-1}(\mathbb{Z})$ into $SL_n(\mathbb{Z})$ by using the extra coefficient to cancel out the determinant.

To embedd $S_n$ in $GL_{n-1}(\mathbb{Z})$ we will consider the permutation matrixes in $GL_n(\mathbb{Z})$. Firstly, we notice that the vector $X \in \mathbb{R}^n$ that is made of only ones is an eigenvector of any permutation matrix.
If $e_1, \dots , e_n$ is the standard basis of $\mathbb{R}^n$, let $B$ be the basis $(X, e_2, \dots , e_n)$ in which a permutation matrix will take this shape:

$ \begin{bmatrix} 1 & * \\ 0 & M(\sigma) \end{bmatrix} $

Where $\sigma$ is a permutation. We may then check that M is an injective morphism from $S_n$ to $GL_{n-1}(\mathbb{Z})$

The answer is yes for $n$ odd, since $S_n$ embeds into $SL_n(\mathbb Z)$ - indeed, for $\sigma\in S_n$ let $P_\sigma$ be the corresponding permutation matrix. Since $\det P_\sigma=\operatorname{sgn}\sigma$ and $n$ is odd, $Q_\sigma:=(\operatorname{sgn}\sigma)\cdot P_\sigma\in SL_n(\mathbb Z)$ and $$Q_\sigma Q_\tau=(\operatorname{sgn}\sigma)(\operatorname{sgn}\tau)\cdot P_\sigma P_\tau=(\operatorname{sgn}\sigma\tau)\cdot P_{\sigma\tau}=Q_{\sigma\tau}.$$ I don't see how it could be generalized to even $n$.