Isomorphic direct product cyclic groups

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Homework Help Overview

The discussion revolves around the isomorphism between cyclic groups, specifically examining whether the cyclic group of order \( p^2 \) is isomorphic to the direct product of two cyclic groups of order \( p \), where \( p \) is a prime number. Participants express confusion regarding the nature of isomorphisms between groups of different structures.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the orders of elements in the groups \( C_{p^2} \) and \( C_p \times C_p \), questioning whether these orders are compatible. There is also a discussion on the nature of isomorphisms between groups with different representations, such as single elements versus ordered pairs.

Discussion Status

Some participants have provided examples of isomorphisms between other cyclic groups, while others are questioning the validity of these examples in the context of the original problem. The discussion is ongoing, with various interpretations being explored without a clear consensus.

Contextual Notes

There is a noted confusion regarding the properties of isomorphisms and the implications of group orders, as well as the use of additive notation in the examples provided. Participants are examining the definitions and assumptions underlying the problem.

blahblah8724
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Help! For p prime I need to show that

C_{p^2} \ncong C_p \times C_p

where C_p is the cyclic group of order p.



But I've realized I don't actually understand how a group with single elements can be isomorphic to a group with ordered pairs!


Any hints to get me started?
 
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blahblah8724 said:
Help! For p prime I need to show that

C_{p^2} \ncong C_p \times C_p

where C_p is the cyclic group of order p.

What possible orders do elements in C_{p^2} have. Are these order compatible with the elements in C_p\times C_p?

But I've realized I don't actually understand how a group with single elements can be isomorphic to a group with ordered pairs!

It certainly is possible. For example, we have C_6\cong C_2\times C_3. Indeed, if we use additive notation (C_6=\{0,1,2,3,4,5\}), then

f:C_6\rightarrow C_2\times C_3

by f(0)=(0,0), f(1)=(1,1), f(2)=(0,2), f(3)=(1,0), f(4)=(0,1), f(5)=(1,2) is an isomorphism.
 
But surely if again we use additive notatoin with C_9=\{0,1,2,3,4,5,6,7,8\} then f:C_9 \to C_3 \times C_3 with
f(0) = (0,0), f(1) = (0,1), f(2) = (0,2), f(3) = (1,0), f(4) = (1,1), f(5) = (1,2), f(6) = (2,0),
f(7) = (2,1), f(8) = (2,2) is as isomorphism?
 
blahblah8724 said:
But surely if again we use additive notatoin with C_9=\{0,1,2,3,4,5,6,7,8\} then f:C_9 \to C_3 \times C_3 with
f(0) = (0,0), f(1) = (0,1), f(2) = (0,2), f(3) = (1,0), f(4) = (1,1), f(5) = (1,2), f(6) = (2,0),
f(7) = (2,1), f(8) = (2,2) is as isomorphism?

No. f(2+2)=f(4)=(1,1). But f(2+2)=f(2)+f(2)=(0,2)+(0,2)=(0,1). So it's not a homomorphism.
 

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