# Cosets of upper triangular matrix

• Nov 24th 2009, 12:43 PM
hfcriske
Cosets of upper triangular matrix
Greets, I'm a bit confused about the definition of cosets as of now, here's my problem:

I'll represent a 3x3 matrix using the first integer for row number and the second for column number. Thus 11 = 1 means there's a 1 in the upper left corner of the matrix, 12 is a step to the right from 11 and 21 is a step down.

Given the group G of 3x3 matrices under multiplication with the following properties:
11 = 1, 12 = a, 13 = b
21 = 0, 22 = 1, 23 = c
31 = 0, 32 = 0, 33 = 1
Where a,b,c are members of Z[n], n positive integer

We have the following subgroup H:
11 = 1, 12 = d, 13 = e
21 = 0, 22 = 1, 23 = -d
31 = 0, 32 = 0, 33 = 1
Where d,e are members of Z[n]

What are the left cosets of H in G?

Any advice/clarification on how to compute this would be great.
• Nov 24th 2009, 01:25 PM
emakarov
I have several questions about the problem statement.

Do you consider regular matrix multiplication, as in linear algebra, but only in Z[n]?

Could you remind why this it a group? E.g., when one has just numbers, i.e., 1x1-matrices, then Z[n] is not always a group under multiplication, is it?

Finally, how is a group determined by a matrix M? Is M a generating element, i.e., does the group consists of all powers of M? Or do we get different group elements by substituting different numbers for a, b, and c?
• Nov 24th 2009, 01:36 PM
hfcriske
I apologize for being a bit vague. Here are some answers:

"Do you consider regular matrix multiplication, as in linear algebra, but only in Z[n]?"

Yes, it's basically regular matrix multiplication in Z[n].

"Could you remind why this it a group? E.g., when one has just numbers, i.e., 1x1-matrices, then Z[n] is not always a group under multiplication, is it?"

A group G with a binary operator * is defined by the following 4 properties:
1: If x is in G and y is in G, then x*y is in G.
2: (x*y)*z = x*(y*z)
3: There is an identity element e in G such that x*e = e*x = x
4: Every x in G has an inverse x' such that x*x' = x'*x = e

"Finally, how is a group determined by a matrix M? Is M a generating element, i.e., does the group consists of all powers of M? Or do we get different group elements by substituting different numbers for a, b, and c?"

Different numbers for a, b and c would likely render different elements, as long as they're different in Z[n]. I suspect generating elements may exist if n is prime, though I'm not sure of this. The group is closed under multiplication, so if M is in G, then so is M^i.
• Nov 24th 2009, 02:42 PM
emakarov
OK, I see now that H is a subset of G because a and c can be arbitrary in G, but in H the corresponding elements are d and -d. I.e., H = {g in G | a = -c}.

I can also see that G is a group because every matrix has an inverse: it's possible to solve the corresponding equations.

It must be the case that $G = H\sqcup g_1H\sqcup\dots\sqcup g_kH$ for some $g_1,\dots,g_k\in G\setminus H$. Here $\sqcup$ is the union of disjoint sets. These $g_iH$ are cosets, and different cosets are disjoint. So the problem is finding those $g_1,\dots,g_k$.

• Nov 24th 2009, 02:54 PM
hfcriske
You've got it 100% correct.

That part about unions of sets you wrote might actually help me already, I'll update here if I get any further today.
• Nov 25th 2009, 12:33 PM
emakarov
Yes, I though that it would be more appropriate to put this problem in the Algebra section.

I have some ideas that need to be checked. Instead of looking directly at $gH$ for various $g$, I used the Fundamental theorem on homomorphisms: If $f:G\to K$ is a homomorphism for some group $K$, then $G/\ker f\cong K$. So if we find $f$ such that $\ker f$ is $H$, then $K$ is (isomorphic) to the group of cosets. To do this, $H$ has to be a normal subgroup, i.e., $gH=Hg$ for all $g\in G$; this can be easily verified.

So, we need to find a homomorphism that maps $H$ into the unit matrix. What is a homomorphism on $G$? For brevity, instead of writing matrices from $G$, I just write triples $(a,b,c)$. The product is defined as follows: $(a,b,c)(a',b',c')=(a+a',b+b'+ac',c+c')$. The subgroup $H$ consists of triples where $c=-a$.

It took me some time to find a homomorphism using "scientific random poking", as they say in Russian. I have $f(a,b,c)=(2(a+c),(a+c)^2,(a+c))$ (needs to be double-checked). I guess one can put 2 in the third component instead of the first. The good thing is that its kernel is indeed $H$.

So, what is the image $K$ of $f$? Well, $a+c$ can be anything, so the image must consist of triples $(2x, x^2, x)$ for all possible $x$. And it seems to me that each such triple produces a coset with $H$. It also seems that indeed every $g\in G$ can be obtained in this way.

I have not thought whether the cosets $(2x, x^2, x)H$ may intersect (i.e., coincide) for different $x$. This may be possible because of modular arithmetic.

Hope this helps. Let us know if there are further difficulties.
• Nov 25th 2009, 02:53 PM
hfcriske
From a first glance of it, it's seems to be exactly what I was looking for.

Many thanks for all your help, this will be very useful to me.