Finitely generated abelian group

Consider the integral matrix R: =

2 2 4 2
4 4 8 5
6 12 12 8
4 10 8 6

Determine the structure of the abelian group given by the generators and relations

AR:= <a1, a2, a3, a4 | R o a = 0 >

And determine the number of elements of order 2 in AR.

Please someone help me out! This wasn't really covered properly in lectures, and I can't seem to find any information on it which makes any sense to me.

Reply 1

around

Hae you covered Smith normal form? Putting your matrix in Smith normal form effectively 'changes the basis' for the finitely generated abelian group given, which should mean you can read off the structure of the abelian group.

Reply 2

Bobifier

For a lack of proofs and notes, I would suggest you look up the smith normal form of a matrix.

Reply 3

Mathlete29

Original post by around

I think we did but it wasn't covered very well.

Is Smith-normal form something to do with row AND column operations? That's where I get a little confused.... with the column operations.

And I don't quite get what the question is asking me to find. Is it just the numbers on the diagonal?

Reply 4

Venomilys

Google smiths normal form of a matrix, it shouldn't take you too long to pick up the concept.

Reply 5

Mathlete29

I have managed to get it into smith normal form. Is the question just asked for the values on the diagonal? Or is it asking for something else?

Reply 6

Jake22

The number of zeros is the rank and the non-zero numbers are the torsion part.

In other words, if your matrix in smith normal form has diagonal entries

x_1,\ldots,x_n

and

m

zeros then your group is isomorphic to

\mathbb{Z}^m\oplus \mathbb{Z}/x_1\mathbb{Z} \oplus \cdots \mathbb{Z}/x_n\mathbb{Z}

.

e.g. if you smith normal form were, say,

\begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}

then your group would be

\mathbb{Z}^3 \oplus \mathbb{Z}/2\mathbb{Z}

The structure theorem for finitely generated modules over PIDS says that ALL fin. generated abelian groups are of this form for some

m

and some

x_1,\ldots,x_n

such that each

x_i

divides

x_{i+1}

.

If you really don't have any of this information in your notes then you can find it easily on the Wikipedia articles:
http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain

and

http://en.wikipedia.org/wiki/Smith_normal_form

where the latter teaches you the technique for converting your matrix to Smith normal form.

If you are just working with abelian groups (rather than modules over a PID) then replace every occurance of 'PID' or 'the principal ideal domain R' etc. with

\mathbb{Z}

.

Once you have written the group in this way, it should be easy to work out how many elements of order two there are.

(edited 11 years ago)

Reply 7

Mathlete29

Actually, I'm still a little confused. When getting it into smith normal form, can I just do row and column operations as I see fit? Or do I need to have those extra matrices either side?

I've searched online but I don't quite get it :s

Reply 8

Jake22

Original post by Mathlete29

Actually, I'm still a little confused. When getting it into smith normal form, can I just do row and column operations as I see fit?

Yes. The whole point of the process is that you have a map

\mathbb{Z}^m \to \mathbb{Z}^n

which you desire to write in a useful form.

This comes from the fact that for a finitely generated abelian group

A

- we have a finite number of generators (represented by the

\mathbb{Z}^n

and a finite number of relations (represented by the

\mathbb{Z}^m

) In fancy language, we may present any fin. gen. abelian group

A

by a short exact sequence

0 \to \mathbb{Z}^m \to \mathbb{Z}^n \to A \to 0

Now, you pick a basis for the domain and codomain and write your map as a matrix. When you perform row and column operations - you are just changing the bases for the domain and codomain. The point being that we want to write the quotient

A

\mathbb{Z}^n

(generators) by the image of

\mathbb{Z}^m

(relations) in a standard format.

Or do I need to have those extra matrices either side?

I don't know what you mean - I assume you are talking about elementary matrices? These are just a way of representing row or column operations by matrix multiplication e.g.

If we take the identity matrix:

I=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}

and switch the rows (i.e. switch the two basis vectors) to get the matrix

A=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}

Then we can write

A = EI

where

E=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}

. Thus performing the row operation of switching the rows is just premultiplying by

E

I've searched online but I don't quite get it :s

Really? There are some great sources in the first few google entries e.g. this at number two in the search for 'Smith Normal Form'

It is probably the case that you need to go back and revise basic linear algebra:

Consider a finite dimensional vector space

V

with (ordered) basis

(x,y,z)

and let

W

be the subspace spanned by,

x+y

and

z

. Then,

W

obviously has the (ordered) basis given by

(x+y,z)

Now consider the linear map

\varphi\colon W \to V

including

W

into

V

as a subspace.

With respect to the bases above, we can represent

\varphi

by the matrix

M_\varphi = \begin{pmatrix} 1 &0 \\ 1 & 0 \\ 0 & 1\end{pmatrix}

. Performing the obvious two row operations gives us the matrix

M' = \begin{pmatrix} 1 &0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}

All we have done is changed the basis of

V

(x+y,z,y)

and written the inclusion

W \to V

as the matrix

M'

with respect to this new basis.

Now, we can write the quotient

V/W

V/W = \langle x+y, z, y \mid x+y, z\rangle

thus we can straight away see that

V/W

is one dimensional with basis

y+W

.

Of course, this is totally banal and pointless - by the rank nullity theorem, we know that

V/W

is one dimensional and we can easily name a basis without putting pen to paper.

The point is that when we generalise this situation to finitely generated abelian groups i.e.

\mathbb{Z}

-modules (which are just 'vector spaces' except we scale by integers in

\mathbb{Z}

rather then elements of a field like

\mathbb{R}

\mathbb{C}

), the situation becomes a tiny bit more complicated and we need to go through that convoluted process above to see what the quotient looks like.

When moving from a field to the integers - we retain a lot of good properties:

i) We can still talk about the 'dimension' of free modules (except now we call it the 'rank') thus if a free module has a basis with say

n

elements - then so do all other bases.

ii) If we take a one dimensional free module (i.e.

\mathbb{Z}

itself) - then every non-zero subspace is spanned by one element.

iii) More generally, every submodule of a free module is free

The main property we lose is that when we quotient a free module by a free submodule - we don't necessarily end up with something free. For example if we take

\mathbb{Z}

which is freely generated by 1 and the subgroup

2\mathbb{Z}

(freely generated by 2) then the quotient

\mathbb{Z}/2\mathbb{Z}

is no longer free but is generated by one element, say

x = 1 + 2\mathbb{Z}

now subject to a relation i.e.

2x = 0

. Thus we have picked up something often called torsion.

In the above example with vector spaces - since vector spaces don't have torsion - when we classify fin. generated. vector spaces - all we need to know is the dimension which is why we could have just used the rank-nullity theorem. Now, with abelian groups, we have to change bases to see what the torsion looks like.

This doesn't happen with vector spaces since everything non-zero in a field is invertible so for example, a non-zero subspace of a one dimensional space

V

is always equal to

V

since it will contain the image of the one of the field. We already saw that this doesn't carry over:

2\mathbb{Z} \neq \mathbb{Z}

since we can't hit the number 2 with any integer to get one.

So what is the effect of this torsion? All it means in terms of the example we did with vector spaces is that we won't always be able to get ones on the diagonal when we perform our row and column operations. If you think about it - the ones just correspond to 0s in the quotient i.e. copies of

\mathbb{Z}/1\mathbb{Z} \cong 0

for each other number on the diagonal, we get finite cyclic summands as well as our free modules (corresponding to the zeros i.e. copies of

\mathbb{Z}

)

Reply 9

Mathlete29

Do the diagonal entries have to be ascending?

I've attempted to put the original matrix into smith normal form and I've ended up with diagonal entries 2, 6, 1..... is that correct?

Reply 10

Jake22

Original post by Mathlete29

Do the diagonal entries have to be ascending?

I've attempted to put the original matrix into smith normal form and I've ended up with diagonal entries 2, 6, 1..... is that correct?

Technically, if you want to get smith normal form (as most people define it), then yes but if you just want to write down your group then of course it doesn't matter. Think about my above and other posts and the links etc. and what the numbers mean.

You are essentially asking what the difference is between

\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z} \oplus 0

and

0 \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z}

On the level of not understanding anything and just manipulating matrices to get answers - are you telling us that you cannot see how you can use row and column operations to transform

\begin{pmatrix}2 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 1\end{pmatrix}

\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6\end{pmatrix}

?

Come on; have you actually really read anything? This is just linear algebra.

If you don't have a good set of notes or a book then just read, for example, this.

All you are doing is changing bases.