Number of elments in sl5

Question

I'm trying to understand something about the lie algebra $sl(5,F)$ when F is a finite field.

Let F5 be a finite field with 5 elements. How many elements exist in $sl(5,F)$ ?

My teacher said there are 120.
my try: there are 24 basis elements in $sl5$ and every element may be multiplied by one of the 5 elements in $F5$. So until now there are 120 elments. Isn't it true that every linear combination of the basis elements will generate new $sl5$ element? And than we will have more than 120 elements in $sl5$?

Answer 1

Was not clear that you meant the lie algebra before you edited - just ignore this solution now

Let us first count the number of elements in $\text{GL}_n(\mathbb{F}_p)$. Recall, that a square matrix is invertible iff all its rows (or columns) are linear independent. For the first row we have $p^n-1$ choices (every vector except the zero vector). To make the second row linear independent form the first one, we can choose any vector excpet for multiples of the first row and thus have $p^n-p$ choices. For the next row we cannot choose a linear combination of the first two and therefore we have $p^n - p^2$ choices etc. This means we get $$\text{ord}(\text{GL}_n(\mathbb{F}_p)) = \prod_{i = 0}^{n-1}p^n - p^i.$$ The group $\text{SL}_n(\mathbb{F}_p)$ is the kernel of the determinant map $\text{det} \colon \text{GL}_n(\mathbb{F}_p) \rightarrow \mathbb{F}_p^{\times}$, which means we have $\text{GL}_n(\mathbb{F}_p)/\text{SL}_n(\mathbb{F}_p) \cong \mathbb{F}_p^{\times}$ by the isomorphism theorem. Thus $$\text{ord}(\text{SL}_n(\mathbb{F}_p)) = \frac{\prod_{i = 0}^{n-1}p^n - p^i}{p-1}.$$ In your case we have $p = n = 5$, which yields very many elements (actually ~$5.6 \cdot 10^{16}$). Already for $p = 5$ and $n=2$ we have $240$ elements.

Answer 2

LATER EDIT: This answer was given for a version of the OP where it was unclear that the main object is a Lie algebra. Please ignore. Some later addition was done to answer the much simpler question, as it is now.

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We will compute the number of elements of $S=\operatorname{SL}(5,F)$ for a field with $q$ elements.

Let us compute first the number of elements of $G=\operatorname{GL}(5,F)$.

• For the first column $(a_1)$, $a_1\in V=F^5$ in a matrix $A$ in $G$ there are $|V|-1=q-1$ possibilities. We fix such a choice.

• For the second column $(a_2)$, $a_2\in V=F^5$ in $A$ there are $|V|-q = q^5-q$ possibilities, since we have to eliminate from the vector field $V$ all $q$ linear combinations of the system (with one element) $(a_1)$. We fix such a further choice.

• For the third column $(a_3)$, $a_3\in V=F^5$ in $A$ there are $|V|-q^2 = q^5-q^2$ possibilities, since we have to eliminate from the vector field $V$ all $q^2$ linear combinations of the system (with two elements) $(a_1,a_2)$. We fix such a further choice.

• For the fourth column $(a_4)$, $a_4\in V=F^5$ in $A$ there are $|V|-q^3 = q^5-q^3$ possibilities, since we have to eliminate from the vector field $V$ all $q^3$ linear combinations of the system (with three elements) $(a_1,a_2, a_3)$. We fix such a further choice.

• For the fifth and last column $(a_5)$, $a_5\in V=F^5$ in $A$ there are $|V|-q^4 = q^5-q^4$ possibilities, since we have to eliminate from the vector field $V$ all $q^4$ linear combinations of the system (with four elements) $(a_1,a_2, a_3, a_4)$.

This way to count gives: $$ |G| = (q^5-q^0) (q^5-q^1) (q^5-q^2) (q^5-q^3) (q^5-q^4)\ . $$ The determinant map from $G$ to $F^\times$ is surjective, the kernel is $S$, so $$ |S|= \frac{|G|}{|F^\times|} = \frac 1{q-1} (q^5-q^0) (q^5-q^1) (q^5-q^2) (q^5-q^3) (q^5-q^4)\ . $$ This is a big number in our case.

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Computer check, here sage

sage: S = SL(5, GF(5))
sage: S.order()
56653740000000000
sage: q = 5
sage: (q^5-1)*(q^5-q)*(q^5-q^2)*(q^5-q^3)*(q^5-q^4) / (q-1)
56653740000000000

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LATER EDIT:

Well, if we have to count matrices of trace zero, then the situation is fairly simple. The way the initial post was organized could not make me see it should go in this direction. Obviously, the linear space $\mathfrak{sl}_5(F)$ of $5\times 5$ matrices of trace zero over a finite field $F$ with $q$ elements sits in the short exact sequence $$ 0 \longrightarrow \mathfrak{sl}_5(F) \longrightarrow M_{5\times 5}(F) \overset{\text{Trace}}\longrightarrow F \longrightarrow 0\ , $$ (or just use the dimension formula without any displayed sequence,) so it has dimension $5^2-1$ and as a vector space of this dimension it has $$ q^{\displaystyle(5^2-1)} $$ elements.

Comment: It is the first time that i here some professor counts the elements of a Lie algebra over a finite field. I am usually dealing with explicit computations for invariants that make at least minimal sense in a given structure. It is OK to ask such a question, please do it, the mathematical truth and a clean desk are important in mathematics, but in this case i cannot omit to make the observation that this compuation is a dead end street. The number obtained does not make any sense, even if we try to work with a computer in that lie algebra elementwise. Especially on this page a dead end should be marked. (The number of elements of a group, in contrast, its order, is a strong invariant, the first one that usually matters.) This is a special case where the comment belongs inside the answer.

Answer 3

We have $\dim_F \mathfrak{sl}_n(F)=n^2-1$. In particular, we have $|F|^{n^2-1}$ elements, since every element $x$ is a unique linear combination of the basis vectors $e_i$, $$ x=a_1e_1+\ldots +a_{n^2-1}e_{n^2-1}. $$ Counting gives $|F|$ to the power of $n^2-1$, and not multiplication $|F|\cdot (n^2-1)$.

For $|F|=5$ and $n=5$ we obtain $$ 5^{5^2-1}=5^{24}=59604644775390625. $$ For $n=2$ we obtain $5^3=125$ elements.