What is the size of the quotient group L/pZ^m?

[Albert01]

Hello,

I have a question that I would like to ask here.

Let $L = \left\{ x \in \mathbb{Z}^m : Ax = 0 \text{ mod } p \right\}$, where $A \in \mathbb{Z}_p^{n \times m}$, $rank(A) = n$, $m \geq n$ and $Ax = 0$ has $p^{m-n}$ solutions, why is then $|L/p\mathbb{Z}^m| = p^{m-n}$?

I am extremely looking forward to your responses!

 

[member38644]

Hello,

The size of the quotient group L/pZ^m is p^(m-n). This is because pZ^m is the subgroup of L that contains all elements of L that are multiples of p. When we take the quotient of L by pZ^m, we are essentially dividing out all of these multiples of p. This leaves us with p^(m-n) distinct cosets, each of which has p^n elements. Therefore, the size of the quotient group is p^(m-n).

In the given scenario, L is the set of all solutions to the equation Ax = 0 mod p. Since there are p^(m-n) solutions to this equation, the size of the quotient group L/pZ^m is also p^(m-n).

I hope this helps to clarify any confusion. Please let me know if you have any further questions. Thank you.