Groups of order p^2 where p is prime

[halvizo1031]

Homework Statement

let p be a prime number and let G be a group with order p^2. the task is to show that G is either cyclic or isomorphic to Zp X Zp.
a. let a, not equal to the identity,be an element in G and A=<a>. What's the order of A.
b. consider the cosets of A: G/A={A,g2A,...gnA}. What's the value of n?

Homework Equations

The Attempt at a Solution

for part a, does the order of A have to do with the index of A in G? If so, how?
for part b, can i use the fact that the index of A divides the order of G? If so, how?

 

[mighty2000]

Hello,

Great question! To answer your first question, yes, the order of A does have to do with the index of A in G. In fact, the order of A is equal to the index of A in G, which can be shown using Lagrange's Theorem. Since the order of G is p^2 and A is a subgroup of G, the order of A must divide the order of G. Therefore, the order of A can only be 1, p, or p^2.

For part b, you are on the right track. You can use Lagrange's Theorem again to show that the index of A in G, denoted by |G:A|, divides the order of G. This means that n, the number of cosets of A in G, must also divide p^2. Since n cannot be 1 (as A is not the only coset in G) and it cannot be p^2 (as there are only p^2 elements in G), the only possible value for n is p. This means that there are p cosets of A in G, which can be written as G/A = {A, g2A, ..., gpA}.

I hope this helps! Let me know if you have any further questions.