# Isomorphic direct product cyclic groups!

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 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?

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 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 $C_6\cong C_2\times C_3$. 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?

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.