If $\gcd(m,n) = 1$ then $\mathbb Z_{mn} \cong \mathbb Z_m \oplus \mathbb Z_m$.

Question

I am familiar with the Chinese Remainder Theorem and I know that it must be used here in some way (Hint given by my lecturer).

All I know so far is $\mathbb Z_{mm} \not\cong \mathbb Z_m \oplus \mathbb Z_m$.

This explains my reaction of 'no way this is true', I tried to play around with $n$ and $m$ using the fact that they're co-prime, but I really do feel like I've hit the wall on this.

(EDIT) Just to elaborate on more work I've done; I played around with small examples (2,3) and now I'm actually really questioning this because how can an isomorphism exist between say $\mathbb Z_{6} \cong \mathbb Z_3 \oplus \mathbb Z_3$? One clearly has 6 elements and the other 9! So how can there be a bijective mapping between these two groups? Perhaps this question has a typo?

Any hints would be appreciated, so please no solutions.

Answer 1

Just a word:

If that two groups be isomorphic so, there are very very similar algebraic patterns between them. For example, if the the left group has four elements of order $6$ so is the right group and vice versa. Here, you can find an elemnt of the right group (since it is cyclic) of ordr $6$ while all elements of the right group has order $3$. This is enough to show that that may be a typo!

Answer 2

If $gcd(m,n) = 1$

then $ (1,1) $ generate the full group $ \mathbb Z_m \oplus \mathbb Z_n$

• If $a(1,1)=(0,0)$, then $a$ necessarily divide both $m$ and $n$

So, $ \mathbb Z_m \oplus \mathbb Z_n$ is a finite group generated by one element and then is isomorphic to $ \mathbb Z_k $ for some $k$. And $ k= order[Z_m \oplus \mathbb Z_n]$