Abstract Algebra: finite groups: subgroups?

Question

Consider the elements
| 0 -1| = A
| 1 0 |
and

| 0 1 | = B
| -1 -1|

from SL(2, R). How do you find |A|, |B|, and |AB|.
Is this answer surprising?

(A and B are 2x2 matrices)

Answer 1

SL(2,R) is the special linear group on 2x2 matrices of real numbers. The determinant of ALL matrices in this group is 1. Since it's a group, AB is also in this group.

That means:

|A| = 1
|B| = 1
|AB| = 1

This answer is not at all surprising.

The answerer Quing Yi G didn't answer your question, and Pascal seems to favor using ambiguous notation. In linear groups, |A| notation is reserved for determinants. That's the reason ord(A) or o(A) is another standard notation. |a| is almost exclusively used in group-theoretic contexts that are not also linear algebraic.

So if you want to talk about the order, why not just look at o(A) and o(B)? o(A) = 4 and o(B) = 3, clearly, but why does AB have infinite order? Does this make sense? Is it surprising?


Anyone can compute determinants or the orders of things. Let's try to actually answer those questions...


SL(2,R) is an infinite group. It's a fairly robust one at that. It does not have a set of generators that generate the group (unlike say a cyclic or dihedral group).

It's not abelian either. That's important. A has finite order, and B has finite order. IF it were abelian (or if A and B commute), we'd know that AB has finite order.

But AB ≠ BA. This means that we cannot necessarily expect them to have finite order. So the answer does make perfectly good sense.

Answer 2

Edit: Okay, judging from the thumbs down, apparently there are some people who don't understand the notation being used here. In group theory, |a| refers to the order of the element a, not the determinant of a. These are two different things. If you don't know what the order of an element of a group is, I refer you to the wikipedia article: http://en.wikipedia.org/wiki/Order_%28group_theory%29 .

Well, finding the order of A and B is as simple as multiplying them out until you get the identity matrix:

[-1, 0] = A²
[0, -1]

[0, 1] = A³
[-1, 0]

[1, 0] = A⁴
[0, 1]

so |A| = 4. Similarly for B:

[-1, -1] = B²
[1, 0]

[1, 0] = B³
[0, 1]

So |B| = 3. Now, to find |AB|, we must first compute AB:

[1, 1] = AB
[0, 1]

Now we note that:

[1, 2] = (AB)²
[0, 1]

[1, 3] = (AB)³
[0, 1]

And in general, given that (AB)^n =
[1, n]
[0, 1]

We know that (AB)^(n+1) =

[1, n+1]
[0, 1]

So by induction, AB^n = [[1, n], [0, 1]] for every n, and in particular, will never equal the identity matrix for any positive n, so |AB|=∞.

As for whether the answer is surprising, I don't know enough about special linear groups to know whether the answer should be surprising.