## Tag Archives: general linear group

### A class of group homomorphisms from CC to GL(n,CC)

Fix an $n \times n$ matrix $M$ over $\mathbb{C}$. Define a mapping $\psi_M : \mathbb{C} \rightarrow \mathsf{GL}_n(\mathbb{C})$ by $\alpha \mapsto \mathsf{exp}(M\alpha)$. Prove that $\psi_M$ is a group homomorphism on $(\mathbb{C},+)$.

Note that $\mathsf{exp}(M\alpha)$ is nonsingular by this previous exercise, so that $\psi_M$ is properly defined.

Now $\psi_M(\alpha + \beta) = \mathsf{exp}(M(\alpha+\beta))$ $= \mathsf{exp}(M\alpha + M\beta)$ $= \mathsf{exp}(M\alpha) \cdot \mathsf{exp}(M\beta)$ by this previous exercise, which equals $\psi_M(\alpha) \psi_M(\beta)$. So $\psi_M$ is a group homomorphism.

### Exhibit a set of representatives of the conjugacy classes of GL(3, ZZ/(2))

Exhibit a set of representatives of the conjugacy classes of $\mathsf{GL}_3(\mathbb{Z}/(2))$.

It suffices for us to give the (essentially unique) rational canonical matrix in each conjugacy class. If $A$ is a $3 \times 3$ matrix over $\mathbb{Z}/(2)$, then the minimal polynomial of $A$ has degree between 1 and 3. There are 14 such polynomials, which we list with their factorizations. (We showed which monic polynomials of degree at most 3 were irreducible in this previous exercise.)

1. $x$ (irreducible)
2. $x+1$ (irreducible)
3. $x^2 = xx$
4. $x^2+1 = (x+1)^2$
5. $x^2+x = x(x+1)$
6. $x^2+x+1$ (irreducible)
7. $x^3 = xxx$
8. $x^3+1 = (x+1)(x^2+x+1)$
9. $x^3+x = x(x^2+1)$
10. $x^3+x+1$ (irreducible)
11. $x^3+x^2 = x^2(x+1)$
12. $x^3+x^2+1$ (irreducible)
13. $x^3+x^2+x = x(x^2+x+1)$
14. $x^3+x^2+x+1 = (x+1)^3$

Note that if $x$ divides the minimal polynomial of $A$, then in particular $A$ is a zero divisor in $\mathsf{Mat}_n(F)$ and so cannot be a unit. So matrices in $\mathsf{GL}_3(\mathbb{Z}/(2))$ cannot have $x$ as a divisor of their minimal polynomial. This eliminates (1), (3), (5), (7), (9), (11), and (13) from our list of candidate minimal polynomials.

An irreducible polynomial of degree 2 is also unsuitable as a minimal polynomial, since the characteristic polynomial of $A$ (having degree 3) cannot divide any power of an irreducible polynomial of degree 2. (Either the characteristic polynomial is irreducible (bad) or has an irreducible linear factor (also bad).) This eliminates (6) from our list of candidate minimal polynomials. The six remaining candidates, once chosen as a minimal polynomial, determine the remaining invariant factors. The possible lists of invariant factors for $A$ are thus as follows.

1. $x+1$, $x+1$, $x+1$
2. $x+1$, $(x+1)^2$
3. $(x+1)(x^2+x+1)$
4. $x^3+x+1$
5. $x^3+x^2+1$
6. $(x+1)^3$

The corresponding rational canonical matrices are as follows.

1. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
2. $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
3. $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$
4. $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
5. $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix}$
6. $\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$

These six matrices are a complete and nonredundant set of representatives of the conjugacy classes of $\mathsf{GL}_3(\mathbb{Z}/(2))$.

### Exhibit the matrices of dimension 2 over QQ having multiplicative order 4

Compute, up to similarity, all the elements of $\mathsf{GL}_2(\mathbb{Q})$ having order 4. Do the same in $\mathsf{GL}_2(\mathbb{C})$.

Our task is to find all the $2 \times 2$ matrices over $\mathbb{Q}$ which satisfy $p(x) = x^4-1$ but not $x^2-1$ or $x-1$. Note that over $\mathbb{Q}$, $p(x) = (x^2+1)(x+1)(x-1)$. If $A$ is a matrix of order 4, then the minimal polynomial must be divisible by $x^2+1$ and must divide $x^4-1$. Since the characteristic polynomial of $A$ has degree 2, there is only one possible list of invariant factors for $A$, namely $x^2+1$. The corresponding rational canonical form matrix is $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Thus every element of $\mathsf{GL}_2(\mathbb{Q})$ of order 4 is similar to this matrix.

Now over $\mathbb{C}$, we have $p(x) = (x+1)(x-1)(x+i)(x-i)$. If $A$ is a matrix of order 4, then its minimal polynomial must be divisible by either $x+i$ or $x-i$, must divide $p(x)$, and must have degree at most 2. There are seven possibilities for the minimal polynomial of $A$, and with the minimal polynomial chosen, the remaining invariant factors are determined (since the characteristic polynomial has degree 2). The possible lists of invariant factors are as follows.

1. $x+i$, $x+i$
2. $(x+i)(x-i)$
3. $(x+i)(x+1)$
4. $(x+i)(x-1)$
5. $x-i$, $x-i$
6. $(x-i)(x+1)$
7. $(x-i)(x-1)$

The corresponding rational canonical forms are as follows.

1. $\begin{bmatrix} -i & 0 \\ 0 & -i \end{bmatrix}$
2. $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
3. $\begin{bmatrix} 0 & -i \\ 1 & -1-i \end{bmatrix}$
4. $\begin{bmatrix} 0 & i \\ 1 & 1-i \end{bmatrix}$
5. $\begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix}$
6. $\begin{bmatrix} 0 & i \\ 1 & i-1 \end{bmatrix}$
7. $\begin{bmatrix} 0 & -i \\ 1 & 1+i \end{bmatrix}$

Every matrix in $\mathsf{GL}_2(\mathbb{C})$ of order 4 is similar (i.e. conjugate) to exactly one matrix in this list.

### Classification of groups of order 2p²

Let $p$ be an odd prime. Prove that every element of order 2 in $GL_2(\mathbb{F}_p)$ is conjugate to a diagonal matrix with $\pm 1$s on the diagonal. Classify the groups of order $2p^2$.

We begin with some lemmas.

Lemma 1: Let $F$ be a field in which $2 \neq 0$. Every element of order 2 in $GL_2(F)$ is conjugate to a diagonal matrix. Proof: Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in GL_2(F)$ have order 2. Now $A^2 = \begin{bmatrix} a^2 + bc & b(a+d) \\ c(a+d) & d^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$; Suppose $a+d \neq 0$. Comparing entries, we see that $b = c = 0$, so that $a^2 = d^2 = 1$. If $a = 1$, then since $a+d \neq 0$, $d = 1$. Similarly, if $a = -1$, then $d = -1$. Thus either $A = \pm I$ or $a+d = 0$. If $A = \pm I$, then $A$ is already diagonal. Suppose $a+d = 0$.

1. If $c = 0$ and $b \neq 0$, let $B = \begin{bmatrix} (a-d)/b & 1 \\ 0 & 1 \end{bmatrix}$. Evidently $B^{-1} = \begin{bmatrix} b/(a-d) & b/(d-a) \\ 0 & 1 \end{bmatrix}$ and $BAB^{-1}$ is diagonal.
2. If $c \neq 0$ and $b = 0$, let $B = \begin{bmatrix} 1 & 0 \\ 1 & (d-a)/c \end{bmatrix}$. Evidently $B^{-1} = \begin{bmatrix} 1 & 0 \\ c/(a-d) & c/(d-a) \end{bmatrix}$ and $BAB^{-1}$ is diagonal.
3. If $c \neq 0$ and $b \neq 0$, let $B = \begin{bmatrix} a/b & 1 \\ 1 & d/c \end{bmatrix}$. Evidently $B^{-1} = \frac{bc}{ad-bc} \begin{bmatrix} d/c & -1 \\ -1 & a/b \end{bmatrix}$ and $BAB^{-1}$ is diagonal. $\square$

Lemma 2: If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in GL_2(F)$ is a diagonal matrix of order 2, then $a = \pm 1$ and $d = \pm 1$. Proof: We have $A^2 = \begin{bmatrix} a^2 & 0 \\ 0 & d^2 \end{bmatrix}$. $\square$

Lemma 3: Let $p$ be an odd prime. Let $m,n \geq 1$ be integers such that for all positive integers $a, b$, $m+a(-1)^n \equiv a+m(-1)^b$ mod $p$. Then $m \equiv 0$ mod $p$ and $n \equiv 0$ mod 2. Proof: If $m \not\equiv 0$ mod $p$, let $a = p$. Then $1 \equiv (-1)^b$ mod $p$ for all $b$, which is absurd. Thus $m \equiv 0$ mod $p$. Now suppose $a \not\equiv 0$ mod $p$; we have $(-1)^n \equiv 1$ mod $p$. Thus $n \equiv 0$ mod 2. $\square$

Now to the main result.

Let $G$ be a group of order $2p^2$, where $p$ is an odd prime. Let $H \leq G$ be a Sylow $p$-subgroup. Since $[G:H] = 2$, $H$ is normal in $G$. Let $K \leq G$ be a Sylow 2-subgroup. Since $H \cap K = 1$, we have $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some group homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying groups of order $2p^2$ is equivalently to finding the distinct semidirect products of this type.

Let $K = Z_2 = \langle x \rangle$.

Let $H = Z_{p^2} = \langle y \rangle$. Now $\mathsf{Aut}(H) \cong Z_{p(p-1)}$; say $\mathsf{Aut}(H) = \langle \alpha \rangle$, where $\alpha(y) = y^2$. Note that $2|p-1$; thus there is a unique element of order 2 in $Z_{p(p-1)}$; namely $\alpha^{p(p-1)/2}$. Thus there are exactly two homomorphisms $\varphi : K \rightarrow \mathsf{Aut}(H)$. $\varphi_1(x) = 1$ yields the semidirect product $H \rtimes_{\varphi_1} K \cong Z_{p^2} \times Z_2$, and $\varphi_2(x) = \alpha^{p(p-1)/2}$ yields the nonabelian group $Z_{p^2} \rtimes_{\varphi_2} Z_2$. These are distinct since one is abelian and one is not.

Now let $H = Z_p^2 = \langle a \rangle \times \langle b \rangle$. Now $\mathsf{Aut}(H) \cong GL_2(\mathbb{F}_p)$. Let $A_1 = I$, $A_2 = -I$, $A_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, and $A_4 = -A_3$. Note that if $\psi : K \rightarrow \mathsf{Aut}(H)$, then $\psi(x)$ has order 1 or 2. If $|\psi(x)|$ is 1 or 2, then $\mathsf{im}\ \psi$ is conjugate to $\langle A_i \rangle$ for some $i$ by Lemmas 1 and 2. By a previous theorem, then, the distinct semidirect products of this form are isomorphic to one of $H \rtimes_{\psi_i} K$. Let $G_1 = H \rtimes_{\psi_1} K$, $G_2 = H \rtimes_{\psi_2} K$, and $G_3 = H \rtimes_{\psi_3} K$. Note that $A_4 = A_2 \cdot A_3$; by a lemma to a previous exercise, $H \rtimes_{\psi_4} K \cong G_3$.

We claim that $G_1$, $G_2$, and $G_3$ are nonisomorphic. Now $G_1 \cong Z_p^2 \times Z_2$ is abelian while the other two are not, so $G_1$ is distinct from $G_2$ and $G_3$.

We will now find presentations for $G_2$ and $G_3$.

Note that $G_2$ is generated by $a$, $b$, and $x$, and $ab = ba$. If we write $a = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, then $xax^{-1} = \psi_2(x)(a)$ $= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \end{bmatrix}$ $= \begin{bmatrix} -1 \\ 0 \end{bmatrix}$ $= a^{-1}$. Similarly, $xbx^{-1} = b^{-1}$. Thus $G_2$ has the presentation $a,b,x \ |\ a^p = b^p = x^2 = 1,$ $ab = ba, xa = a^{-1}x,$ $xb = b^{-1}x \rangle$.

$G_3$ is also generated by $a$, $b$, and $x$, and $ab = ba$. Similarly, $xax^{-1} = a$ and $xbx^{-1} = b^{-1}$. Thus $G_3$ has the presentation $\langle a,b,x \ |\ a^p = b^b = x^2,$ $ab = ba, xa = ax, xb = b^{-1}x \rangle$.

We will now compute the center of $G_2$. We can see that every element of $G_2$ has the form $a^ib^jx^k$ where $0 < i,j \leq p$ and $0 < k \leq 2$. There are $2p^2$ such forms, and we know by other means that $|G_2| = 2p^2$. Thus every element can be written in this form uniquely. Let $s = a^mb^nx^\ell \in Z(G_2)$ and $a^ib^jx^k \in G_2$; using several previous lemmas, we have that $a^{m+i(-1)^\ell}b^{n+j(-1)^\ell}x^{\ell+k}$ is equal to $a^{i+m(-1)^k}b^{j+n(-1)^k}x^{\ell+k}$. Comparing exponents and using Lemma 3, we have $m \equiv n \equiv 0$ mod $p$ and $\ell \equiv 0$ mod 2; thus $s = 1$, and we have $Z(G_2) = 1$.

Note that $Z(G_3)$ is nontrivial since $a \in Z(G_3)$. In particular, $G_2 \not\cong G_3$.

Thus there are five distinct groups of order $2p^2$, $p$ an odd prime.

### Exhibit the set of all invertible diagonal matrices with diagonal entries equal as a direct product

Let $G = \{ [a_{i,j}] \in GL_n(F) \ |\ a_{i,j} = 0\ \mathrm{if}\ i > j, a_{i,i} = a_{j,j}\ \mathrm{for\ all}\ i,j \}$, where $F$ is a field, be the group of upper triangular matrices all of whose diagonal entries are equal. Prove that $G \cong D \times U$, where $D$ is the group of nonzero multiples of the identity matrix and $U$ the group of strictly upper triangular matrices.

Recall that $\overline{SL}_n(F)$ is normal in $GL_n(F)$, where $\overline{SL}_n(F)$ denotes the upper triangular matrices with only 1 on the diagonal. Then $U = G \cap \overline{SL}_n(F)$ is normal in $G$. Now if $A \in D$, we have $A = kI$ for some nonzero field element $k$. Then for all $M \in GL_n(F)$, we have $AM = kIM = kM = Mk = MkI = MA$, so that $D \leq Z(GL_n(F))$; more specifically, $D$ is normal in $G$.

We can see that $D \cap U$ is trivial, since every element in $U$ has only 1s on the main diagonal. Finally, if $A \in G$ has the element $k$ on the main diagonal, then $k^{-1}A \in U$, and in fact $A = kI \cdot k^{-1}A \in DU$. Thus $G = DU$; by the recognition theorem, we have $G \cong D \times U$.

### Exhibit Sym(n) as a subgroup of a general linear group

Let $G_i = F$ be a field for all $i$ and use the preceding exercise to show that the set of $n \times n$ matrices over $F$ such that each row and column contains exactly one 1 and all other entries are 0 is a subgroup of $GL_n(F)$ isomorphic to $S_n$. (These matrices are called permutation matrices since they simply permute the standard basis $e_1,\ldots,e_n$ (as above) of the $n$-dimensional vector space $F^n$.)

We know that if $F$ is a finite field then $\mathsf{Aut}(F^n) \cong GL_n(F)$. This isomorphism $\zeta$ can be defined as follows: given $\theta \in \mathsf{Aut}(F^n)$, $\zeta(\theta)$ is the matrix in $GL_n(F)$ whose $i$th row is precisely $\theta(e_i)$. (In particular, $\zeta$ is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In the previous exercises, we found an injective group homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(F^n)$. Combining these results we have an injective group homomorphism $\Psi : S_n \rightarrow GL_n(F)$, computed as follows: $\Psi(\pi) = [e_{\pi(1)}\ \cdots\ e_{\pi(n)}]^T$. We can see that each $\Psi(\pi)$ is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus $S_n$ is identified with a subgroup of $GL_n(F)$ consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.

Thus we have proven the result for a finite field $F$; we began with this case because it was shown previously that $\mathsf{Aut}(F^n) \cong GL_n(F)$. If this is true for arbitrary fields then the same proof carries over to arbitrary fields $F$; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields $F$ by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not $k = 0$ in $F$ for some integer $k$ never arises, so that all computations hold for arbitrary fields (in which $1 \neq 0$) and in fact the set of permutation matrices over any $F$ is closed under matrix multiplication.

### Compute the number of Sylow p-subgroups in a given general linear group

Prove that the number of Sylow $p$-subgroups of $GL_2(F_p)$ is $p+1$.

We know that $|GL_2(F_p)| = (p^2-1)(p^2-p) = p(p-1)^2(p+1)$. By Sylow’s Theorem, the number $n_p$ of Sylow $p$-subgroups divides $(p-1)^2(p+1)$ and is congruent to 1 mod $p$.

We saw in the previous exercise that the strictly upper triangular matrices $\overline{UT}_2(F_p)$ form a Sylow $p$-subgroup. We now show that $UT_2(F_p) \leq N_{GL_2(F_p)}(\overline{UT}_2(F_p))$.

Let $A = \begin{bmatrix} a & b \\ 0 & c \end{bmatrix}$ be an invertible upper triangular matrix; then $A^{-1} = \begin{bmatrix} 1/a & -b/ac \\ 0 & 1/c \end{bmatrix}$. Now if $B = \begin{bmatrix} 1 & d \\ 0 & 1 \end{bmatrix}$ is strictly upper triangular, we have $ABA^{-1} = \begin{bmatrix} 1 & ad/c \\ 0 & 1 \end{bmatrix}$, which is strictly upper triangular. Thus the normalizer of $\overline{UT}_2(F_p)$ contains the upper triangular matrices in $GL_2(F_p)$. There are $p(p-1)^2$ of these, since diagonal entries are arbitrary nonzero field elements, while the upper right entry is unrestricted. Now $[G : UT_2(F_p)] = [G : N_G(\overline{UT}_2(F_p))][N_G(\overline{UT}_2(F_p)) : UT_2(F_p)]$. By Sylow’s Theorem, then, $n_p$ divides $p+1$. Since $n_p$ is congruent to 1 mod $p$, either $n_p = 1$ or $n_p = p+1$; however, $n_p \neq 1$ since the strictly upper and strictly lower triangular matrices are distinct Sylow $p$-subgroups of $GL_2(F_p)$.

### The strictly upper triangular matrices form a Sylow subgroup in the general linear group over ZZ/(p)

Show that the subgroup $\overline{UT}_n(F_p)$ of strictly upper triangular matrices (i.e. with only 1 in diagonal entries) in $GL_n(F_p)$ is a Sylow $p$-subgroup.

We know that the order of $GL_n(F_p)$ is $\prod_{i=0}^{n-1} (p^n - p^i) = \prod_{i=0}^{n-1} p^i(p^{n-i} - 1)$ $= (\prod_{i=0}^{n-1} p^i)(\prod_{i=0}^{n-1}(p^{n-i}-1))$. Note that $p$ does not divide $p^{n-i}-1$ for any $i$. Moreover, we have $\prod_{i=0}^{n-1} p^i = p^{\sum_{i=0}^{n-1} i} = p^{n(n-1)/2}$. Thus (by definition) a Sylow $p$-subgroup of $GL_n(F_p)$ has $p^{n(n-1)/2}$ elements.

Recall that a matrix is strictly upper triangular if all diagonal entries are 1, all lower entries are 0, and all upper entries can be any field element. Then the number of elements in $\overline{UT}_n(F_p)$ is $p^M$, where $M$ is the number of entries above the main diagonal. We can see that this number is $M = \sum_{i=1}^n n-i = \sum_{i=0}^{n-1} i = n(n-1)/2$. Thus $\overline{UT}_n(F_p)$ is a Sylow $p$-subgroup in $GL_n(F_p)$.

### The general linear group over the complex numbers is a union of conjugates of a proper subgroup

Let $G = GL_2(\mathbb{C})$ and let $H = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \ |\ a,b,c \in \mathbb{C}, ac \neq 0 \right\}$. Prove that every element of $G$ is conjugate to some element of $H$. Deduce that $G$ is the union of conjugates of $H$. [Hint: Show that every element of $G$ has an eigenvector.]

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in G$; then $ad-bc \neq 0$.

• If $c = 0$, then $IAI^{-1} = A \in H$.
• If $b = 0$, let $B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$; moreover, $B^{-1} = B$. Then $BAB^{-1} = \begin{bmatrix} d & c \\ b & a \end{bmatrix} \in H$.
• If $b,c \neq 0$ and $a = d$, let $B = \begin{bmatrix} c/\sqrt{c} & 0 \\ 1/\sqrt{b} & 1/\sqrt{c} \end{bmatrix}$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$. Evidently $B^{-1} = \begin{bmatrix} 1/\sqrt{c} & 0 \\ -1/\sqrt{b} & c/\sqrt{c} \end{bmatrix}$. Note that $BAB^{-1} = \begin{bmatrix} a-\sqrt{bc} & bc \\ 0 & d+\sqrt{bc} \end{bmatrix} \in H$.
• Suppose now that $b,c \neq 0$ and $a \neq d$. Let $B = \begin{bmatrix} 0 & a-d \\ 1/(d-a) & \omega \end{bmatrix}$, where $\omega = \left(1 + \sqrt{1 + 4bc/(d-a)^2}\right)/2c$. Note that $\mathsf{det}(B) = 1$, so that $B \in G$. Evidently $B^{-1} = \begin{bmatrix} \omega & d-a \\ -1/(d-a) & 0 \end{bmatrix}$. Moreover, we see that $BAB^{-1} = \begin{bmatrix} \omega c(a-d) + d & -c(a-d)^2 \\ 0 & \omega c(d-a) + a \end{bmatrix} \in H$.

Thus every element of $G$ is conjugate to some element of $H$; clearly then $G = \bigcup_{g \in G} gHg^{-1}$.

### Compute the index of the special linear group in the general linear group

Let $F$ be a finite field of order $q$ and let $n$ be a positive integer. Prove that $[GL_n(F):SL_n(F)] = q-1$.

We proved in a previous exercise that $GL_n(F)/SL_n(F) \cong F^\times$, where $F^\times$ denotes the set of all elements in $F$ which have a multiplicative inverse. By definition, this is all elements except 0. So the index is $q-1$.