Show that $SL_2(\Bbb{F}_4)\cong A_5$.

Question

The Problem: Show that $SL_2(\Bbb{F}_4)\cong A_5$.

Source: Abstract Algebra, $\mathit{3^{3d}}$ edition by Dummit and Foote.

My Attempt: Sufficient to show that $G:=SL_2(\Bbb{F}_4)$ is simple.

Approach #1: Since $|G|=60$, it is therefore sufficient to show that $|Syl_5(G)|>1$.

But I don't quite know how $SL_2(\Bbb{F}_4)$ actually looks like, therefore I could not produce the two distinct Sylow 5-subgroups needed.

Approach #2: BWOC, suppose $\exists H\lhd G$ such that $1<H<G$. Note $|H|=2, 3, 4, 5, 6, 10, 12, 15, 20$ or $30$, thus $|G/H|=30, 20, 15, 12, 10, 6, 5, 4, 3$ or $2$. But this did not really yield a contradiction either.

Approach #3: Find an isomorphism $\varphi: G\to A_5$. This seems mission impossible.

I am now dead-stuck. Any help would be greatly appreciated.

Answer 1

Here is one standard way.

You have the natural module $V = \mathbb{F}_4^2$ for $G = SL_2(\mathbb{F}_4)$. Then $G$ acts on the set $$X = \{ \langle v \rangle : v \in V, v \neq 0\}$$ of lines in $V$. (In other words, $X$ is the projective space $\mathbb{P}(V)$).

Steps:

• Check that $|X| = 5$. Let $\varphi: G \rightarrow S_5$ be the homomorphism corresponding to the action of $G$ on $X$.
• Check that the kernel of $\varphi$ is trivial.
• Thus $G$ is isomorphic to $\varphi(G)$.
• Since $A_5$ is the only subgroup of order $60$ in $S_5$, we have $\varphi(G) = A_5$.
• Conclude $G \cong A_5$.