TITLE: Can any finite group of order $n$ be embedded in $SL_n(\mathbb{Z})$?
QUESTION [6 upvotes]: If $G$ is a finite group of order $n$, it can be embedded in $S_n$ by Cayley's theorem and therefore in $GL_n(\mathbb{Z})$ and $SL_{n+1}(\mathbb{Z})$. Can $G$ be embedded in $SL_n(\mathbb{Z})$?

REPLY [3 votes]: 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 $GL_{n-1}(\mathbb{Z})$