How to find [G:H]?

Question

Let $F$$=GF(11)$ be finite field of 11 elements. G is group of all non-singular n$\times$n matrices over F.$H$ is subgroup of those matrices whose determinant is 1. Then $[G:H]$=?

Answer 1

Interesting question will be, if you will ask yourself and calculate orders of $G$ & $H$. then obviously you will also have the $[G:H]$

Answer 2

Hint 1: Consider the homomorphism of groups $\varphi:G\to F^*$ which takes a matrix to its determinant.

Hint 2: What do we know about $\varphi$? Is it surjective? What is its kernel?

Answer 3

I find it intriguing to break the discussion of this problem into several stages, moving from groups in general first on to general matrix groups over arbitrary fields $\Bbb F$ second and then finally narrowing the focus to the case of finite fields, closing with the case $\Bbb F = \Bbb Z_{11}$.

From general group theoretic considerations, we have, for any group $G$, that two elements $x, y \in G$ are in the same (left) coset of a subgroup $H$ if and only if $x^{-1}y \in H$:  if $x, y \in aH$, then $x = ah_1$ and $y = ah_2$ for some $h_1, h_2 \in H$; thus

$x^{-1}y = (ah_1)^{-1}ah_2 = h_1^{-1}a^{-1}ah_2 = h_1^{-1}h_2 \in H; \tag{1}$

likewise if $x^{-1}y \in H$, then $x^{-1}y = h \in H$, so $y = xh$, whence $yH = xhH = xH$, since $hH = H$; the similar result (fairly) obviously holds for right cosets $Ha$, by essentially the same reasoning.

If we apply this idea to the matrix group $G = GL(n, \Bbb F)$, that is, to the group of $n \times n$ invertible matrices with entries taken from the (arbitrary) field $\Bbb F$, we see that $M_1, M_2 \in GL(n, \Bbb F)$ are in the same coset of $H = SL(n, \Bbb F)$, that is, the subgroup of $GL(n, \Bbb F)$ of matrices of determinant $1$, if and only if $M_1^{-1}M_2 \in SL(n, \Bbb F)$ is unimodular; that is, $\det(M_1^{-1}M_2) = 1$.  But then, since $\det$ is a multiplicative function on matrices,

$\det(M_1^{-1}M_2) = \det(M_1^{-1})\det(M_2) = (\det(M_1))^{-1} \det(M_2); \tag{2}$

here we have used $\det(M_1^{-1}) = (\det(M_1))^{-1}$ which also follows from the multiplicative property, since

$\det(M_1^{-1})\det(M_1) = \det(M_1^{-1}M_1) = \det(I) = 1. \tag{3}$

We thus have $(\det(M_1))^{-1} \det(M_2) = 1$ or $\det(M_1) = \det(M_2)$.  If follows then that $M_1$ and $M_2$ are in the same coset of $SL(n, \Bbb F)$ precisely when $\det(M_1) = \det(M_2)$. These facts imply that $\det$ is an invariant of the cosets of $SL(n, \Bbb F)$ in $GL(n, \Bbb F)$, in the sense that $\det$ is constant on each such coset, and that the map

$\det: GL(n, \Bbb F) / SL(n, \Bbb F) \to {\Bbb F}^\ast \tag{4}$

from the set of cosets of $SL(n, \Bbb F)$ to $\Bbb F^\ast$ is both well defined and injective. However, it is easy to see that $\det$ is surjective as well: for any nonzero $a \in \Bbb F$, consider the matrix in $GL(n, \Bbb F)$ formed by taking the identity matrix $I$ and replacing any single $1$ on the diagonal with $a$; it is easy to see that the resulting matrix $A$ is nonsingular, hence $A \in GL(n, \Bbb F)$, and that $\det A = a$; then for any $M \in SL(n, \Bbb F)$ we have $\det(AM) = \det(A) \det (M) = \det(A) = a$. Thus $\det$ maps $GL(n, \Bbb F)/SL(n, \Bbb F)$ onto $\Bbb F^\ast$, and it then follows that the cosets of $SL(n, \Bbb F)$ in $G = GL(n, \Bbb F)$ are in one-to-one correspondence with the elements of $\Bbb F^\ast$. Indeed, since $\det: GL(n, \Bbb F) \to \Bbb F^\ast$ is multiplicative, it is in fact a group homomorphism: $\det(M_1 M_2) = \det (M_1)\det(M_2)$. Thus, since $\ker \det = SL(n, \Bbb F)$ we have that $GL(n, \Bbb F) / SL(n, \Bbb F) \cong \Bbb F^\ast$, and so $\text{card} (GL(n, \Bbb F) / SL(n, \Bbb F)) = \text{card} (\Bbb F^\ast)$.

In the event that $\Bbb F$ is finite, then so is $GL(n, \Bbb F)$, and so $[GL(n,\Bbb F):SL(n, \Bbb F)]$, the index of $SL(n, \Bbb F)$ in $GL(n,\Bbb F)$, i.e., the number of cosets of $SL(n, \Bbb F)$, is well-defined. Since $[GL(n, \Bbb F):SL(n, \Bbb F)]= \text{card} (GL(n, \Bbb F) / SL(n, \Bbb F))$, we find that

$[GL(n, \Bbb F):SL(n, \Bbb F)] = \text{card}(\Bbb F^\ast) = \text{card}(\Bbb F) - 1. \tag{5}$

Finally, choosing $\Bbb F = \Bbb Z_{11}$, we have $\text{card}(\Bbb F^\ast) = 10$; thus $[G:H] = [GL(n, \Bbb F):SL(n, \Bbb F)] = 10$.

The above demonstration may of course be synopsized via the observation that $\det$, being multiplicative, is in fact a homomorphism from $G$ onto $\Bbb F^\ast$, and that $\ker \det = H$, whence $\det: G/H \to \Bbb F^\ast$ is an isomorphism, whence, again if $\Bbb F$ is finite, $[G:H] = \text{card} (\Bbb F^\ast) -1$. This, in essence, is the argument suggested by Jared in his answer.

Hope this helps. Cheers,

and of course,

Fiat Lux!!!