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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.

Bernard
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Florian Suess
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  • ?? Why $\mathbb Z_3 \oplus \mathbb Z_3$? You wrote $(2,3)$ just above. – John Brevik Sep 18 '18 at 06:28
  • Since 2 and 3 are co-prime, then if the statement is true then indeed $\mathbb Z_{6} \cong \mathbb Z_3 \oplus \mathbb Z_3$ – Florian Suess Sep 18 '18 at 06:29
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    BTW, the Chinese Remainder Theorem is the whole thing -- so be sure you really are familiar with it! – John Brevik Sep 18 '18 at 06:29
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    Oh, I see. Sorry. Yes, there's a typo. Should be $\mathbb Z_m \oplus \mathbb Z_n$. Otherwise it wouldn't make much sense :) – John Brevik Sep 18 '18 at 06:31
  • Mhmm, well then it follows directly from CRT, what a weird question he has given me. So my direct (EDIT) conclusion shows correct intuition as to why it is incorrect (from one perspective?) – Florian Suess Sep 18 '18 at 06:33
  • I don't quite follow the last part. I do understand the first part, though. It's still a useful exercise, as CRT is generally stated in terms of elementary number theory and this is an interesting and useful illustration of how the same statement expressed in the language of modern algebra looks a bit different. – John Brevik Sep 18 '18 at 06:36

2 Answers2

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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!

Mikasa
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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]$

Slepecky Mamut
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