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Groups and Theory: Smith Normal Form

I have done the first part of the question but now I'm struggling on the second part. How do I start this and what does the question want? Thank you for your time (the answer for the first part is below)

Unparseable latex formula:

\begin{bmatrix}[br]1 & 0 & 0 & 0 \\[br]0 & 3 & 0 & 0 \\[br]0 & 0 & 21 & 0 \\[br]0 & 0 & 0 & 0 [br]\end{bmatrix}

Reply 1

Mark13

Original post by gardash

Unparseable latex formula:

\begin{bmatrix}[br]1 & 0 & 0 & 0 \\[br]0 & 3 & 0 & 0 \\[br]0 & 0 & 21 & 0 \\[br]0 & 0 & 0 & 0 [br]\end{bmatrix}

I'm not familiar with the definition of "isomorphism type", but I would assume it means write your group as the direct product of cyclic groups of prime order.

Reply 2

gardash

Original post by Mark13

I'm not familiar with the definition of "isomorphism type", but I would assume it means write your group as the direct product of cyclic groups of prime order.

Is it

Z_{1}*Z_{3}*Z_{3}*Z_{7}

Reply 3

Mark13

Original post by gardash

Is it

Z_{1}*Z_{3}*Z_{3}*Z_{7}

Almost - think about what the zero in the diagonal means:

You've got a relation matrix R for some generators a1, ..., a4, and you've done some operations to it to get a new relation matrix R' for a new set of generators b1, ... , b4. So the group is generated by the elements b1, ..., b4, subject to the relations

R' b = 0

where b is the column vector with entries b1, ..., b4. From the form of R', we get that b1=0, 3 b2= 0, 21 b3= 0, and 0 b4 = 0. You corretly identified that the subgroup generated by b1 is trivial, the subgroup generated by b2 is isomorphic to C3 and the subgroup generated by b3 is isomorphic to C21 (which you correctly simplified to C3 x C7). What is the subgroup generated by b4 isomorphic to?

Reply 4

gardash

I am not sure, could it possibly be just Z again?

Reply 5

Mark13

Original post by gardash

I am not sure, could it possibly be just Z again?

Yep that's right - there is no relation on b4 other than that it commutes with the other generators, so the subgroup generated by it must be isomorphic to Z.

Reply 6

gardash

Original post by Mark13

Yep that's right - there is no relation on b4 other than that it commutes with the other generators, so the subgroup generated by it must be isomorphic to Z.

Thank you very much for your help sir

Reply 7

Mark85

Here is one way to think about Smith Normal Form and what is going on here (especially if you have done the more general case of fg modules over PIDs).

Think about what would happen if you replaced

\mathbb{Z}

throughout with a field, e.g.

\mathbb{C}

.

If you think about the procedure for Smith Normal Form - you would always end up with a diagonal matrix consisting of a certain number of 1s followed by some zeros. Indeed, what you would be working out would be the quotient of a vector space given by identifying the solutions of a linear system. The Smith Normal Form procedure is then just a change of basis where you write a basis for that solution space and complete it so that the remaining basis vectors span the quotient.

Smith normal form over a PID is only a slight generalisation of this - because not every element is invertible - fg modules aren't always free but they do have a basis and the only extra phenomenon is the torsion coefficients that appear due to lack of invertibility but aside from that - they are as close as possible to finite dimensional vector spaces as you can get.

Original post by Mark13

I'm not familiar with the definition of "isomorphism type", but I would assume it means write your group as the direct product of cyclic groups of prime order.

Isomorphism type means what you think it means - fg abelian groups are totally classified upto isomorphism and there is a canonical representative for each isomorphism class. However, in this instance - I would have to assume that they were looking for the invariant factor decomposition i.e.