Proving Subgroup Criteria for Matrix Groups

In summary, the conversation discusses how to show that a set of integer matrices with specific conditions forms a subgroup. The first criterion is showing that the set is non-empty, which is proven by providing an example of a matrix in the set. The second criterion is showing closure under the binary operation of GL2R, which can be proven by multiplying two matrices and showing that the resulting matrix is also in the set. The third criterion is showing that the inverse of each element in the set also belongs to the set. This can be easily proven using the theorem that states if a subset has the property that a, b belong to the set, then ab^-1 also belongs to the set.
  • #1
Firepanda
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0
http://img145.imageshack.us/img145/9528/matrixex6.jpg

I have 3 subgroup criterion.

For the first i have to show H is non empty, well here the matrix

2 1
5 3

is an element of H, so it is none empty.

The second I have to show that H is closed under the binary operation of GL2R. How do I do this? By definition ig g, h are elements of H, then gh is an element of H. I have no idea how to show this is true for it?

I can find another matrix that would be in the group of H, and I can show the product of the matrices is 1, but do I have to prove it? And how?

Thirdly I have to show the inverse of each element of H belongs to H, which is easy.
 
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  • #2
Your set of matrices are all integer matrices with the lower left entry divisible by 5 and determinant 1. Dealing with the determinant 1 condition is pretty easy if you remember det(gh)=det(g)*det(h). Let g1=[[a1,b1],[5c1,d1]] and g2=[[a2,b2],[5c2,d2]]. To show closure multiply g1*g2 and show the lower left corner is still divisible by 5 and all the entries are integers. For inverses compute (g1)^(-1) and show it has the same properties.
 
  • #3
I think the fastest way might be to exploit that theorem that says that if it's a subset (which it is) that has the property that [itex]a,b \in H \Rightarrow ab^{-1} \in H[/itex] then H is a subgroup. I say that because just exploiting the fact that the determinant is a homomorphism, it's not hard to show that ab^{-1} is in H without doing a lot of math.
 

Related to Proving Subgroup Criteria for Matrix Groups

1. What are subgroups?

Subgroups are a type of mathematical structure that consist of a subset of elements from a larger group that also form a group when operated on by the same operation as the larger group.

2. How do you determine if a subset is a subgroup?

To determine if a subset is a subgroup, you must check if it satisfies the following three conditions: 1) closure under the operation, meaning that when any two elements in the subset are operated on, the result must also be in the subset, 2) associativity, meaning that the order in which operations are performed does not matter, and 3) the existence of an identity element, meaning there is an element in the subset that when operated on with any other element in the subset, results in the other element.

3. What is the order of a subgroup?

The order of a subgroup is the number of elements in the subgroup. It can be the same as the order of the larger group or a smaller number, depending on the size of the subset.

4. How are subgroups related to matrices?

Subgroups can be formed from matrices when they are operated on using matrix multiplication. The subset of matrices that satisfy the three conditions for being a subgroup will form a subgroup.

5. How can matrices be used to solve homework problems related to subgroups?

Matrices can be used to represent and manipulate elements in a subgroup, making it easier to solve homework problems related to subgroups. They can also be used to verify if a subset is a subgroup by checking if the three conditions for being a subgroup are satisfied.

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