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> [!definition] Definition. ([[projective special linear group]])
> Let $\mathbb{F}_{m}$ be finite [[field]] of $n$ elements.
> The **projective special linear g[](field.md)md)imension $n$ over $\mathbb{F}_{m}$** is the [[quotient group|quotient]] of the corresponding [[special linear group]] by $\{ \pm I \}$: $\text{PSL}_{n}(\mathbb{F}_{m}):=\text{SL}_{n}(\mathbb{F}_{m}) / \{ \pm I \}.$
> [!basicexample] Example. ($\text{PSL}_{2}(\mathbb{F}_{5}^{}) \cong A_{5},$)
>
>
Set $G=\text{PSL}_{2}(\mathbb{F}_{5})$. Let $\text{GL}_{n}$ the [[general linear group]]. Recall that $|\text{GL}_{2}(\mathbb{F}_{5})|=(5^{2}-1)(5^{2}-5)$, by [[general linear group over a prime field]] Now, [[determinant|det]]$: \text{GL}_{2}(\mathbb{F}_{5}) \to \mathbb{F}_{5}^{\times}$ is a [[surjection|surjective]] [[group homomorphism|homomorphism]] whose [[kernel]] is $\text{SL}_{2}(\mathbb{F}_{5})$. By the [[first isomorphism theorem]], then, we have $\text{GL}_{2}(\mathbb{F}_{5}) / \text{SL}_{2}(\mathbb{F}_{5}) \cong \mathbb{F}_{5}^{\times}$. So $|\text{SL}_{2}(\mathbb{F}_{5})|=\frac{|\text{GL}_{2}(\mathbb{F}_{m})|}{|\mathbb{F}_{5}^{\times}|}=120.$
[[order of quotient group is quotient of orders|Conclude that]] $|\text{PSL}_{2}(\mathbb{F}_{5})|=60$.
>
Next, recall from our proof that [[A5 is the unique simple group of order 60]] that if a [[group]] has $6$ $5$-[[p-Sylow subgroup|sylow subgroups]] it must be $A_{5}$. We know that $G$ has $n_5=1$ or $n_{6}=6$ (order 5 subgroups) by [[the Sylow theorems]]. The [[matrix]] $\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}$
[[generating set of a group|generates]] a [[subgroup]] of [[order of a group|order]] $5$ in $G$, as does the [[matrix]]
>
$\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}$
but the [[subgroup]]s are not equal. Thus $n_{6}=6$ and we conclude that $\text{PSL}_{2}(\mathbb{F}_{5}^{}) \cong A_{5},$
where $A_{5}$ denotes the [[alternating group]] on $5$ letters.
^8f522c
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#### References
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