# Ring: Proof of Group

1. Aug 5, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
Let R be the ring of all 2*2 matrices, over Zp, p a prime. Let G be the set of elements x in the ring R such that det x ≠ 0. Prove that G is a group.

2. Relevant equations
Matrix is invertible in ring R.

3. The attempt at a solution
Group properties and ring properties are similar I think.
Group and Ring - closure, associativity, identity (zero in Rings?)
Is this how I am supposed to approach the problem? By proving the common properties of a group and ring?

2. Aug 5, 2014

### jbunniii

$G$ will be a group under multiplication, not addition. So you can ignore the addition and address the following questions:

1. What is the multiplicative identity? Is it contained in $G$?
2. Is matrix multiplication associative?
3. Is the set of matrices with nonzero determinant closed under multiplication and inverses?

Hint for part 3: can you prove and apply the identities $\det(AB) = \det(A)\det(B)$ and $\det(I) = 1$?

3. Aug 5, 2014

### Staff: Mentor

The elements of G already belong to a ring, so these elements already satisfy all of the properties of a ring (which must include an addition operation and a multiplication operation). If you list the properties that a group must satisfy, this should be pretty easy to prove.

Groups are much more basic than rings.

4. Aug 5, 2014

### Justabeginner

1. Multiplicative identity in a ring is 1, right? But 1 is not in G, because G consists of the set of matrices for which ad - bc ≠ 0, but (1) is not a 2*2 matrix.
2. Yes; [(ab)c]ij = (ab)ik ckj = (ailblk)ckj = ail(blkckj) = ail(bc)lj = [a(bc)]ij
3. Yes, for A= (a b c d), and B= (e f g h), it was proven that det(A)det(B)=det(AB), and det(I)= 1 by 1/ad-bc (ad-bc 0 0 ad-bc) since 1/ad-bc where ad-bc≠0 is the inverse of matrix M. The set is closed.

5. Aug 5, 2014

### jbunniii

We often use the notation "1" for the multiplicative identity in a ring, but it doesn't have to literally mean the number 1. In this case, the ring consists of 2x2 matrices over $Z_p$, so if there is an identity, it must be a 2x2 matrix over $Z_p$. What is the definition of "identity"?

Seems like there are some $\sum$ symbols missing?
Where did you use the fact that the matrix elements are from $Z_p$ (where $p$ is prime), as opposed to $Z_n$ for some nonprime integer? Also, can you explain in more detail how you concluded that the set is closed? How do the determinant formulas prove this?

6. Aug 5, 2014

### Justabeginner

It means 1*n = n. The identity should be (1 0 0 1).

Sorry, I completely forgot to insert them.

I failed to mention that the elements are from Zp, p being a prime. Zp is a field if and only if p is prime, and the properties hold true because of this.
Since the determinant is nonzero, and the inverse has been solved for: Det(M)* I = 1, the set is closed under multiplication and inverses. A group must contain an element and its inverse, which this set of matrices does.

7. Aug 5, 2014

### jbunniii

This equation doesn't make any sense. If I is the identity matrix, then the left hand side is a matrix. But the right hand side is (apparently) the scalar 1.
Why? Suppose that $A \in G$. You need to explain in detail why $A^{-1}$ exists, and why $A^{-1} \in G$.

Similarly, suppose that $A,B \in G$. Why is $AB \in G$?

8. Aug 5, 2014

### Justabeginner

M * M-1 = I. The original matrix times its inverse is equal to the Identity Matrix.

If an element exists, then its inverse must exist because the identity element is always in a group (definition of group).
Groups are always closed under products, as well.

9. Aug 5, 2014

### jbunniii

1. How do you know that $M$ has an inverse?
2. How do you know that $\det(M^{-1}) \neq 0$?
But your goal is to prove that $G$ is a group. You are effectively arguing that "it's a group because it's a group."

10. Aug 5, 2014

### Justabeginner

M has an inverse because M= (a b c d) has another matrix such that M * (other matrix) = Identity Matrix. We know that the det(M-1) ≠ 0 because det(M) ≠0, and 1/det(M)≠0, ever. This ring is specifically an integral domain, I believe.

The property of a ring is closure under multiplication and identity exists.
So since the ring is a subgroup of the M(2*2) then it is a group itself.

11. Aug 5, 2014

### jbunniii

How do you know this is true? It would not be true if the matrix elements were from $Z_n$ where $n$ is not prime.
OK, so you're using $\det(M)\det(M^{-1}) = \det(M M^{-1}) = \det(I) = 1$. You need to state this more explicitly.
An integral domain is commutative by definition. Matrix multiplication is not commutative.
What ring is a subgroup of $M_{2\times 2}$? I'm not following your logic. You need to explain why $G$ is closed under multiplication. If $A$ and $B$ are two matrices with nonzero determinant, what can you say about $\det(AB)$?

12. Aug 5, 2014

### Justabeginner

If Det(A) ≠0 and Det(B)≠0, then Det(AB)≠0 too.

13. Aug 5, 2014

### jbunniii

...because $\det(AB) = \det(A)\det(B)$. And how do you know that if $\det(A)$ and $\det(B)$ are nonzero, then so is $\det(A)\det(B)$? Keep in mind that we are working in $Z_p$, not the real numbers, so you have to justify what may seem like standard facts. Again, this would not be true in $Z_n$ for nonprime $n$.

14. Aug 5, 2014

### Justabeginner

For A, B in Zp, where p is prime, Det(A)Det(B) = Det(AB).

15. Aug 6, 2014

### pasmith

No, it's true for every matrix ring that $\det(AB) = \det(A)\det(B)$. The question here is: if $\det(A)$ and $\det(B)$ are both non-zero, is it necessarily the case that $\det(A)\det(B)$ is non-zero?

This is equivalent to showing that, for $a \in \mathbb{Z}_p$ and $b \in \mathbb{Z}_p$, if $a \neq 0$ and $b \neq 0$ then $ab \neq 0$.

This isn't true for general $\mathbb{Z}_n$: for example in $\mathbb{Z}_{10}$ we have $2 \times 5 = 0$. So what's special about $\mathbb{Z}_p$ for prime $p$?

EDIT: You may also at some point want to make use of the fact that in any ring of 2x2 matrices, $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} ad - bc & 0 \\ 0 & ad - bc \end{pmatrix}$$ and $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc.$$

Last edited: Aug 6, 2014
16. Aug 8, 2014

### Justabeginner

Is G, technically the non-abelian group GL(2, R)? I approached this problem in that manner.