Help with Subgroups of GL(n,R): JL's Questions Answered

  • Thread starter Thread starter jeff1evesque
  • Start date Start date
jeff1evesque
Messages
312
Reaction score
0
I've chosen another problem in the book, but am not clear of how to begin. Could someone help me.

Instructions:
Determine whether the given set of invertible nxn matrices with real number entries is a subgroup of GL(n, R).

Problem:
The nxn matrices with determinant -1.

Definition:
If a subset H of a group G is closed under the binary operation of G and if H with the induced operation from G is itself a group, then H is a subgroup of G.

Questions:
In the instructions (above), what does the notation GL(n, R) denote? I looked for similar notations in the respective section of the book, but couldn't find anything.

What is meant by the following: ...and if H with the induced operation from G is itself a group..."?

Is the binary operation defined as the equation for taking the determinant?

To test whether we have a subgroup, do we just use the group axioms: closure, associativity, identity element, inverse element? I'm guessing if the group axioms pass (for our given subset), then the set is a subgroup?

Thanks,JL
 
Physics news on Phys.org


GL(n,R) is the general linear group of nxn matrices over the real numbers. also known as the invertible nxn real matrices.

H with the induced operation from G is just technical stuff... G is a group with a multiplication operation, say *. Now when you ask if H, a subset of G, is a group, you have to pick which operation you're going to use. Of course you're only really interested in *, but * is a function over G, not H. So you defined the induced operation say *H to be * restricted to only H in its domain. No thinking is actually required for this, and often people will just not mention it (since it's obvious which operation you want)

So in this case, your group operation is going to be matrix multiplication, and the induced operation is still matrix multiplication.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top