Does alternating group $A_5$ is a subgroup of $GL(4,\mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, \mathbb Z)$?
In case $GL(4, \mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?
I want to know how many subgroups are there in $GL(4, \Bbb Z)$ isomorphic to $A_5$. Is there a unique one?
Thank you so much in advance.
For any $n$, $S_n$ embeds in $GL(n-1,\Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,\ldots,v_n$ where $v_1,\ldots,v_{n-1}$ are the standard unit vectors in $\Bbb Z^{n-1}$ and $v_n=-v_1-\cdots-v_{n-1}$. By definition $v_1+\cdots+v_n=0$. For each permutation $\tau\in S_n$, there's a unique endomorphism of $\Bbb Z^{n-1}$ taking each $v_k$ to $v_{\tau(k)}$ (this works due to $v_1+\cdots+v_n=0$). So $S_n$ embeds into $\text{Aut}( \Bbb Z^{n-1}) =GL(n-1,\Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,\Bbb Z)$.