# Modulo groups

1. Jun 15, 2008

### noisysignal

hi,
i am very new to group theory .... just started reading infact. I could really appreciate some guidance on the following problem with which i am stuck.

Let G be the set of all 2x2 matrices [a b ; c d] (ad-bc != 0) where a,b,c,d are all integers modulo p ( a prime number). Multiplication operation is defined as in the case of Matrices with the understanding that addition and multiplication of entries will be modulo p.
(i) To verify that G (with the operation defined) forms a finite non-abelian group
(ii) Given a prime p, to find the order of group G.

I have been able to verify without much difficulty, the following :
*) Closure under multiplication
*) Existence of identity element
*) Associativity under product

With reference to the existence of inverse, i am not able to prove that the inverse element is actually part of the set. Am i missing some observation/fact ? That the group is finite and non-abelian is easy enough once the inverse thing is shown. The second part of the question requires to separate those elements from the p^4 matrices which do not satisfy the |A| != 0 condition.

2. Jun 15, 2008

### matt grime

The inverse of an invertible 2x2 matrix is the same irrespective of whether the entries are in F_p (as yours are), or anything else. Have you met matrices before? If so just use that formula. Even if not, you can work out the inverse by hand. With the above notation, give [a,b;c,d] what conditions must [u,v;x,y] satisfy for

[a,b;c,d].[u,v;x,y] =[1,0;0,1]?

I don't understand why you need inverses to prove it is finite (there are only a finite number of 2x2 matrices in F_p, p^4 as you point out). Non-abelianness is just by example, so I don't see why the inverse thing is necessary there.

The last part of the question is linear algebra - it's the same over any field (don't worry what a field is).

For a matrix to be invertible, the rows must be linearly independent. In how many ways can I choose the first row so that the first row is not a linearly dependent vector on its own? Now, how many ways can I choose the second row so that both vectors are linearly independent?

3. Jun 15, 2008

### HallsofIvy

Staff Emeritus
The inverse of a matrix 2 by 2 matrix is a 2 by 2 matrix isn't it? And, since A*A-1= A-1*A= I" isn't the "inverse of A-1 obvious? And therefore the fact that it has an inverse? What more do you need?

4. Jun 15, 2008

### noisysignal

hi, thanks very much for your replies. I am sorry for not framing my query in a clearer fashion. The existence of an inverse element in the set is a necessary condition for it to be called a group. For a matrix A = [a b; c d] the inverse (as has been correctly pointed out) will be A-1 = 1/|A|* Adjoint(A). Now the problem is that i am not able to convince my self, that the individual elements of A-1 will be integers modulo p ( as is a condition on the set G). I am trying to verify/prove this result.

Thanks for helping

5. Jun 15, 2008

### matt grime

By definition they will be. Since ad-bc is not zero, mod p, it has a multiplicative inverse - you will surely have proved that sest {1,2,..,p-1} is a group under multiplication mod p. So 1/det(A) is defined, as are the entries of Adj(A).

The entries in A are better thought of as a set of representatives of the residue classes of integers mod p. Call the [0],[1],...,[p-1] if that helps until you have that set in your mind.

Last edited: Jun 15, 2008
6. Jun 15, 2008

### maverick280857

I think what Kartik wants to prove explicitly is that the entries of $A^{-1}$ are integers, for every $A \in G$. He and I incorrectly computed $A^{-1}$ arriving at a matrix with non-integer entries, which cannot be an element of $G$.

Also, $adj(A)$ has integer entries. But when you divide every element of $adj(A)$ by $det(A)$, you should get a matrix with integer entries. With p = 7 and $A = [2, 3 ; 4, 5]$, the inverse of $A$ has non-integer entries. So I guess we're making a mistake somewhere.

That the existence of the inverse in G should follow from the definition is appealing, but not immediately obvious.

7. Jun 16, 2008

### matt grime

Sorry? Don't get you. We aren't looking at matrices with integer entries, we're looking at matrices over the field F_p with p elements. The entries are therefore equivalence classes of integers.

8. Jun 17, 2008

### maverick280857

Ah ok, thanks Matt..that makes sense now. I misinterpreted the problem as involving matrices with integer elements from the set {0, 1, ..., p-1}.

Kartik, here is something you might want to check out: http://en.wikipedia.org/wiki/Equivalence_class.

The following is closely related.

[QUOTE = "Wikipedia entry (http://en.wikipedia.org/wiki/Equivalence_class)"]
The rational numbers can be constructed as the set of equivalence classes of ordered pairs of integers $(a,b)$ with $b$ not zero, where the equivalence relation is defined by

$(a,b) ~ (c,d)$ if and only if $ad = bc$

Here the equivalence class of the pair $(a,b)$ can be identified with rational number $a/b$.
[/QUOTE]

Last edited: Jun 17, 2008