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I'm trying to understand why how you can determine whether two groups of the form $\mathbb{Z}_n$ are isomorphic to each other. More specifically why is $\mathbb{Z}_2 \oplus \mathbb{Z}_3 \cong \mathbb{Z}_6$ but $\mathbb{Z}_3 \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_{15}$?

Is there some method that I'm missing of determining that two groups are isomorphic?

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(i) What you say you are trying to understand and your "more specifically" are not equivalent. (ii) $\mathbb{Z}_3\oplus\mathbb{Z}_5$ is isomorphic to $\mathbb{Z}_{15}$; what do you mean when you claim it is not? –  Arturo Magidin Apr 25 '12 at 21:20
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Your examples are instances of the Chinese remainder theorem. en.wikipedia.org/wiki/Chinese_remainder_theorem –  Fredrik Meyer Apr 25 '12 at 21:24
    
The title does not reflect the question. –  lhf Apr 25 '12 at 21:34
    
Two canonical forms are provided by the fundamental theorem of finitely generated abelian groups. –  lhf Apr 25 '12 at 21:36
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2 Answers 2

Point 1. What you are asking is not what you claim to be trying to understand.

Point 2. $\mathbb{Z}_3\oplus\mathbb{Z}_5$ is isomorphic to $\mathbb{Z}_{15}$. What makes you say it isn't?

Point 3. Perhaps you want to compare with $\mathbb{Z}_3\oplus\mathbb{Z}_6$, which is not isomorphic to $\mathbb{Z}_{3\times 6}=\mathbb{Z}_{18}$.

The reason is that while $2$ and $3$ are relatively prime, $3$ and $6$ are not. Every element of $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has order that divides $\mathrm{lcm}(a,b)$. If $\mathrm{lcm}(a,b)\lt ab$, then the group cannot be cyclic, because a finite group of order $k$ is cyclic if and only if there is (at least) one element of order exactly $k$. If every element has order dividing (and hence, less than or equal) to something strictly smaller than the order of the group, then the group cannot be cyclic.

Since $\mathrm{lcm}(a,b)=ab$ if and only if $a$ and $b$ are relatively prime, this tells you that a necessary condition for $\mathbb{Z}_a\oplus\mathbb{Z}_b$ to be cyclic is for $\gcd(a,b)=1$. In order to show that it is enough (sufficient), then notice that the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$: $$\begin{align*} k(1,1) = (0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b&\iff (k,k)=(0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b\\ &\iff k\equiv 0\pmod{a}\text{ and }k\equiv 0\pmod{b}\\ &\iff a\text{ divides }k\text{ and }b\text{ divides }k\\ &\iff \mathrm{lcm}(a,b)\text{ divides }k. \end{align*}$$ So the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$. Hence, if $\mathrm{lcm}(a,b)=ab$, then $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has (at least) one element of order $ab$, and so it is cyclic.

As a consequence, we have that in (finite) direct sum of cyclic groups, $$\mathbb{Z}_{a_1}\oplus\cdots\mathbb{Z}_{a_k}$$ is cyclic, and isomorphic to $$\mathbb{Z}_{a_1\times\cdots\times a_k}$$ if and only if the $a_i$ are pairwise relatively prime.

Point 4. The question you say you are trying to understand (When two groups of the form $\mathbb{Z}_n$ are isomophic) is: $\mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_m$ if and only if $n=m$. They have to have the same size to be isomorphic; and if they are the same size and both cyclic, then they are isomorphic.

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What do you mean when you say 'pairwise relatively prime'? Do you mean for example $\mathbb{Z}_6 \oplus \mathbb{Z}_7 \oplus \mathbb{Z}_{12} \cong \mathbb{Z}_{504}$ because $6,7$ are coprime and $7,12$ are coprime? –  user26069 Apr 26 '12 at 9:09
    
@user26069: No. "pairwise relatively prime" means "every pair is relatively prime". So what you give doesn't work because although 6 and 7 are relatively prime, and 7 and 12 are relatively prime, 6 and 12 are not. –  Arturo Magidin Apr 26 '12 at 14:44
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What characterizes the group $\mathbb{Z}_n$ is that

1) It has $n$ elements;

2) it is cyclic.

If you consider $\mathbb{Z}_2\oplus\mathbb{Z}_3$, then you can easily check that $1\oplus 1$ is a generator. Same with $\mathbb{Z}_3\oplus\mathbb{Z}_5$. But if you consider the group $\mathbb{Z}_2\oplus\mathbb{Z}_2$, for example, then every element has order 2, so it cannot be cyclic: it is a group of order 4, but it is not isomorphic to $\mathbb{Z}_4$.

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