0

Exercise :

Given the following groups : $$\mathbb Z_{3^2} \times \mathbb Z_{5^2} \cong \mathbb Z_{225} $$ $$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{5^2} \cong \mathbb Z_3 \times \mathbb Z_{75}$$ $$\mathbb Z_{3^2} \times \mathbb Z_{5} \times \mathbb Z_{5} \cong \mathbb Z_{45} \times \mathbb Z_5$$ $$\mathbb Z_{3} \times \mathbb Z_{3} \times \mathbb Z_{5} \times \mathbb Z_{5} \cong \mathbb Z_{15} \times \mathbb Z_{15}$$ show that $2$ of them are not isomorphic.

Attempt :

For the first of two that I have been asked to find, one can see that :

$$\mathbb Z_3 \times \mathbb Z_3 \times \mathbb Z_{5^2} \ncong \mathbb Z_3 \times \mathbb Z_3 \times \mathbb Z_5 \times \mathbb Z_5$$ since if this was an isomorphism, it's clear that the $\mathbb Z_3 \times \mathbb Z_3$ parts are isomorphic, which leads that it also should be $\mathbb Z_{5^2} \cong \mathbb Z_5 \times \mathbb Z_5 $. This is not possible though, as this would imply that the group $\mathbb Z_5 \times \mathbb Z_5$ is cyclic, which is false.

Similar goes for the case :

$$\mathbb Z_{3^2} \times \mathbb Z_5 \times \mathbb Z_5 \ncong \mathbb Z_3 \times \mathbb Z_3 \times \mathbb Z_5 \times \mathbb Z_5$$

Is the above approach correct and complete ? Could it be phrased better if so ?

Rebellos
  • 21,324
  • How do you prove $\mathbf Z_5\times\mathbf Z_5$ is not cyclic? Does it result from a theorem you've learnt? – Bernard Jan 21 '18 at 14:06
  • @Bernard Yes ! In order for the group $\mathbb Z_m \times \mathbb Z_n$ to be cyclic, it would need to be $\gcd(m,n) = 1$ which means that $m,n$ should be coprime. In that case though, $\gcd(5,5) = 5$ so it's not cyclic. – Rebellos Jan 21 '18 at 14:09
  • But are you sure this is a necessary condition (it's true, but I wonder on which basis you're allowed to say so). – Bernard Jan 21 '18 at 14:14
  • @Bernard I'm actually studying for an exam with past questions so it would be nice to learn how should I phrase such parts. On an answered question of our teacher similar to this, the answer did not involve a justification but that could be just for the sake of time ! – Rebellos Jan 21 '18 at 14:15
  • 1
    Actually, the simplest way is to show one of the rings has an element of a certain order and the other ring has not. For instance, the cyclic group $\mathbf Z_{225}$ has an element of order $225$, but none of the other has. – Bernard Jan 21 '18 at 14:22
  • @Bernard What of them are rings ? We haven't been introduced to rings in this semester (we'll do so at Algebra II and III). I got the element stuff though, thanks! – Rebellos Jan 21 '18 at 14:31
  • 1
    OK, you don't need to use this fact anyway. Just see them as groups. – Bernard Jan 21 '18 at 14:47
  • @Bernard Perfect, understood ! – Rebellos Jan 21 '18 at 14:49

1 Answers1

1

$\mathbb Z_{225}=\mathbb Z_{15^2}$ is not isomorphic to $\mathbb Z_{15} \times \mathbb Z_{15}$ since $(15, 15)\neq1$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005