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I'm quite new to group theory, so apologies if this is a silly question to ask: it just interested me that $$\require{cancel}\Bbb Z_{4}\ncong\Bbb Z_2\oplus \Bbb Z_2$$ whereas $$\Bbb Z_{6}\cong \Bbb Z_2\oplus \Bbb Z_3$$ For the first non-equation (I don't know the correct terminology to use here), since $\Bbb Z_2$ is a cyclic group of order $2$ it follows that each element in $\Bbb Z_2\oplus \Bbb Z_2$ has at most order $2$ and thus the isomorphism is impossible. But for the second equation, I cannot tell whether it's true at glance. I only know that the orders of elements in both LHS and RHS are $1,2,3,6$. I do not know whether RHS is cyclic till I write it explicitly, which is quite inefficient.

So I am wondering if there is a better way to handle this kind of problem in general: is there a know-it-at-first-glance method to tell whether $$\Bbb Z_{mn}\cong \Bbb Z_{m}\oplus \Bbb Z_{n}?$$ Or even more generally, what about $$\Bbb Z_{m_1m_2\cdots m_n}\cong \Bbb Z_{m_1}\oplus \Bbb Z_{m_2}\oplus\cdots\oplus\Bbb Z_{m_n}?$$

Christoph
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Vim
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    https://en.wikipedia.org/wiki/Chinese_remainder_theorem – User3773 Oct 21 '15 at 12:10
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    Please, note that the syntax \Bbb Z has been deprecated for 20 years in LaTeX and \mathbb{Z} is to be preferred. – egreg Oct 21 '15 at 12:21
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    No need to apologize: no legitimate question is silly. – lhf Oct 21 '15 at 12:22
  • @egreg Thanks. I don't know this yet. I simply chose the one with less typesetting trouble.. – Vim Oct 21 '15 at 12:24
  • @lhf Thank you sir. And thanks for your nice explanation as well. – Vim Oct 21 '15 at 12:26
  • @Vim If you plan to use LaTeX for producing your documents, it's better changing habits, so your input is valid on both LaTeX and MathJax. It's a nuisance that Math.SE doesn't accept \[...\] instead of $$...$$, though. – egreg Oct 21 '15 at 12:26
  • @egreg My LaTeX seems to accept \Bbb quite well.. I have not yet seen it cause any problems. – Vim Oct 21 '15 at 12:29

3 Answers3

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Yes, there is a nice rule that says $$ \mathbb{Z}_{n}\oplus \mathbb{Z}_m \simeq \mathbb{Z}_{mn} $$ if and only if $\gcd(m,n) = 1$.

Also, the external direct product $\oplus$ is associative, so $A\oplus (B\oplus C) \simeq (A\oplus B) \oplus C$. So the result extends to the more general case that you consider.

Thomas
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  • It's associative up to isomorphism, not literally associative. – Matt Samuel Oct 21 '15 at 12:47
  • @MattSamuel: Right, so as I write $A\oplus (B\oplus C) \simeq (A\oplus B) \oplus C$. It is common to identify groups that are isomorphic. – Thomas Oct 21 '15 at 13:00
  • Ever since I was awakened to the fact that isomorphic things aren't necessarily exactly the same I've been picky about this. It was a very harmful misunderstanding when I took algebraic topology. Anyway, I suppose I can forgive you this time :P. – Matt Samuel Oct 21 '15 at 13:03
  • @MattSamuel: Yeah, when I teach Abstract Algebra I usually make a big deal about the differences between being isomorphic and equal. But I then soften up and we start saying things like: "There is only one group of order one". Likewise, I usually say that the (external) direct product is associative. – Thomas Oct 21 '15 at 17:45
  • Why the downvote? – Thomas Dec 11 '15 at 00:46
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Consider the mapping $$ \varphi\colon\mathbb{Z}\to\mathbb{Z}_m\oplus\mathbb{Z}_n $$ defined by $$ \varphi\colon x\mapsto (x+m\mathbb{Z},x+n\mathbb{Z}) $$ which is a group homomorphism.

Its kernel is $$ \ker\varphi=\{x\in\mathbb{Z}:x\in m\mathbb{Z}, x\in n\mathbb{Z}\} =m\mathbb{Z}\cap n\mathbb{Z}=k\mathbb{Z} $$ where $k$ is the lowest common multiple of $m$ and $n$.

Thus $\varphi$ induces an injective homomorphism $$ \tilde\varphi\colon \mathbb{Z}/\ker\varphi=\mathbb{Z}_k \to\mathbb{Z}_m\oplus\mathbb{Z}_n $$ which is surjective if and only if $k=mn$, by looking at the number of elements in the domain and codomain, that is, $\gcd(m,n)=1$.

Conversely, it's easy to see that $\mathbb{Z}_m\oplus\mathbb{Z}_n$ is not cyclic if $\gcd(m,n)>1$, so it can't be isomorphic to $\mathbb{Z}_{mn}$.

Generalize to any (finite) number of factors.

egreg
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Note that $\Bbb Z_m\oplus \Bbb Z_n$ has order $mn$ and exponent $lcm(m,n)$ (that is, every element has order at most $lcm(m,n)$).

Therefore, $\Bbb Z_m\oplus \Bbb Z_n$ cannot be cyclic unless $lcm(m,n)=mn$, and this is equivalent to $gcd(m,n)=1$.

Thus, if $gcd(m,n)>1$, then $\Bbb Z_{mn}\not\cong \Bbb Z_m\oplus \Bbb Z_n$.

The converse is also true, and is the subject of the Chinese Remainder Theorem.

lhf
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