Isomorphic direct product cyclic groups!

  • #1
Help! For p prime I need to show that

[itex] C_{p^2} \ncong C_p \times C_p [/itex]

where [itex] C_p [/itex] is the cyclic group of order p.



But I've realised 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?
 

Answers and Replies

  • #2
22,129
3,297
Help! For p prime I need to show that

[itex] C_{p^2} \ncong C_p \times C_p [/itex]

where [itex] C_p [/itex] is the cyclic group of order p.

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

But I've realised 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 [itex]C_6\cong C_2\times C_3[/itex]. Indeed, if we use additive notation ([itex]C_6=\{0,1,2,3,4,5\}[/itex]), then

[itex]f:C_6\rightarrow C_2\times C_3[/itex]

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.
 
  • #3
But surely if again we use additive notatoin with [itex] C_9=\{0,1,2,3,4,5,6,7,8\} [/itex] then [itex] f:C_9 \to C_3 \times C_3 [/itex] 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?
 
  • #4
22,129
3,297
But surely if again we use additive notatoin with [itex] C_9=\{0,1,2,3,4,5,6,7,8\} [/itex] then [itex] f:C_9 \to C_3 \times C_3 [/itex] 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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