Proof: $\mathbb{Z}_2 \oplus \mathbb{Z}_6 \not \cong \mathbb{Z}_{12}$

Question

I have the following question :

Proof : $\mathbb{Z}_2 \oplus \mathbb{Z}_6 \not \cong \mathbb{Z}_{12}$

The problem is that I think that $\mathbb{Z}_2 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{12}$. I don't understand why is the following not an injective and surjective function:

$$f:\mathbb{Z}_{12} \rightarrow \mathbb{Z}_2 \oplus \mathbb{Z}_6$$ for all $a\in \mathbb{Z}_{12}$ so that $f(a)=(a \mod 2,a \mod 6)$.

I'm not really sure how to approach such questions, in general when asked proof/disproof $\mathbb{Z}_x \oplus \mathbb{Z}_y \cong \mathbb{Z}_{xy}$, Any tips? Any ideas?

Thank you!

Answer 1

Following your approach we have $ f (11\mod 12)=f (5\mod 12)=(1\mod 2, 5\mod 6) $, hence $ f$ is not injective.

Btw, the following general theorem is true: $\Bbb {Z}_{mn}$ is isomorphic to $\Bbb {Z}_m\oplus \Bbb {Z}_n $ iff $\gcd (m,n)=1$. You can find a proof of this result here.

Answer 2

$\mathbb Z_6\oplus \mathbb Z_{2}$ has no element of order $12$, whereas $\mathbb Z_{12}$ has one.

Answer 3

HINT: consider an arbitrary element $(a, b)\in\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}/12\mathbb{Z}$. What can you say about $6(a, b)$ (that is, $(a, b)$ added to itself $6$ times)? Is that true of every element of $\mathbb{Z}/12\mathbb{Z}$?

Answer 4

Exactly one of those groups has exactly one element of order $2$.

Answer 5

The other answers correctly prove that the two groups aren't isomorphic.

I'll add enough to help you find out why your proof isn't one: look at $f(2)$ and $f(8)$, or $f(0)$ and $f(6)$. Then you should be able to see what your $f$ actually does.