# How do find all cosets?

Can someone explain to me how to find all the cosets of a set like H={A in GL(n) | det(A) = 1} in GL(n) (set of invertible n x n matrices)?

It's obvious how to find all the cosets for something simple like 3Z (set of all multiples of 3) in Z, we just find elements in Z, but not in 3Z that partitions Z, namely choosing 0, 1, 2 as those elements in Z gives the cosets 0+3Z, 1+3Z, 2+3Z, which partitions Z.

It's also obvious in the case where the group is finite. But how will I find it for H in GL(n) above? I'm not sure how to choose elements g in GL(n) but not in H, such that all the cosets of the form gH will partition GL(n). Any suggestions?

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You could use diagonal n x n matrices where the product of the diagonal entries is any integer you wish.

This way you know that you have a determinant that ranges through all integers, which would give you all the elements in GL(n) but of course you'd ignore det = 1

I'm not sure if this is as precise as you want though.

Can someone explain to me how to find all the cosets of a set like H={A in GL(n) | det(A) = 1} in GL(n) (set of invertible n x n matrices)?

It's obvious how to find all the cosets for something simple like 3Z (set of all multiples of 3) in Z, we just find elements in Z, but not in 3Z that partitions Z, namely choosing 0, 1, 2 as those elements in Z gives the cosets 0+3Z, 1+3Z, 2+3Z, which partitions Z.

It's also obvious in the case where the group is finite. But how will I find it for H in GL(n) above? I'm not sure how to choose elements g in GL(n) but not in H, such that all the cosets of the form gH will partition GL(n). Any suggestions?

When searching for cosets of infinite subgroups, it helps to come up with some sort of common property of all the elements in the subgroup H. In your example involving 3Z, look at the representatives of each coset. 0+3Z represents all integers whose remainder is 0 when divided by 3. So you want to think about a property that any particular element (and hence all the elements) in your coset might have. Notice that all the elements of H have determinant 1, which might suggest that the key is to look at the determinant of an invertible matrix. I think ocohen had the right idea.

GL(n) is the collection of all invertible n x n matrices. Let's just play around with a few cosets (actually, just one). Let B be an n x n matrix whose determinant is det(B) = b. Then what is BH?

$$BH = \{ BA \mid A \in H \}$$

and what is det(BA) for any A in H? det(BA) = det(B) x det(A) = det(B) = b. So BH contains a family of invertible n x n matrices whose determinant is b. This raises the question: does BH contain ALL matrices with determinant b?

So let's just pick some random, invertible matrix B' whose determinant is also hapens to be det(B') = b? Can you prove that BH actually contains B' as well? To show that B' is also a member of H, we simply need to prove that B' = BA' for some A' in H. Well, try A' = B^{-1} B'. You need to prove that A' is indeed a member of H (this should be easy to do).

Notice that

$$BA' = BB^{-1}B' = I_n B' = B$$

So the coset BH actually contains ALL matrices with determinant b. So it seems to me that every coset is precisely determined by the determinant of the elements of that coset.

Hello all new to the physics forum. I have stayed up many nights working out this problem and looking for answers I hope someone here might help me please. I understand the definitions and even the proofs but for the life of me I am havaing problems. I'm finding left and right cosets but I am failing to see how to arrive at the final answer. My book has the following for a left coset:

Let G = {S(3), O } Let H = {e, (1,2)}

(1,3,2) {(1,3,2)e, (1,3,2)(1,2)} = {(1,3,2), (2,3)} is just one of the left cosets......

Please oh please how do they get (1,3,2)(1,2) = (2,3) I am probably missing something very simple. Thanks all

I don't understand the two lines preceding the last line - there may be typos or misprints.

The last line makes sense so long as the mappings within the cycles proceed from left to right but the cycles themselves are evaluated right to left.

That is if $\phi=(1,3,2)$ is defined as

$\phi:1\mapsto 3$
$\phi:3\mapsto 2$
$\phi:2\mapsto 1$

$\psi=(1,2)$ is defined as

$\psi:1\mapsto 2$
$\psi:2\mapsto 1$

and $(1,3,2)(1,2)$ is defined as the mapping $\phi\circ \psi$ defined in turn by $\phi\circ \psi:x\mapsto \phi(\psi(x))$.

Then
$\phi\circ \psi:1\mapsto \phi(\psi(1))=\phi(2)=1$
$\phi\circ \psi:2\mapsto \phi(\psi(2))=\phi(1)=3$
$\phi\circ \psi:3\mapsto \phi(\psi(3))=\phi(3)=2$

So $(1,3,2)(1,2)$ is then $(2,3)$.

Some authors use the convention that permutations are evaluated from left to right, in which case the last line would be incorrect. However the line above suggests that this may be the convention in use, but you have lost a permutation $(1,3,2)$ from the left. In which case the LHS should be $(1,3,2)(1,3,2)(1,2)$.

If this is evaluated left to right you get:

$1\rightarrow 3\rightarrow 2\rightarrow 1$
$2\rightarrow 1\rightarrow 3\rightarrow 3$
$3\rightarrow 2\rightarrow 1\rightarrow 2$

again giving $(2,3)$.

Thank you Martin so much I realize that my book at one point list cycles two different ways!

Let a= {1,3,2) and b = {1,2}

1--a(2)--3
2--a(1)--2
3--a(3)--1

so the answer was (1,3) i did not understand why they did not write (1,3)(2) as they had done in the other section on composites of permutations.

Again thank you

P.S. if interested my book is An Introduction to Abstract Algebra with Notes to the Future Teacher by Olympia E. Nicodemi, Melissa A. Sutherland, Gary W. Towsley

Cycles with a single element are usually dropped for convenience. This also allows the calculations to be fairly ambiguous about which $S_n$ is in use. The above calculations would apply equally well to $S_4, S_5, \dots S_\omega, \dots$.