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I am currently reading the book Group Theory in a Nutshell for Physicists by Anthony Zee and he wrote something about isomorphism I could not quite understand.

Now we come in for a bit of a surprise. Consider $\Bbb{Z}_2 \otimes \Bbb{Z}_3$ consisting of $(n, m)$, which we compose by $(n + n' \mod 2, m + m' \mod3)$. Again, we start with $(0, 0)$ and add $(1, 1)$ repeatedly:

$(0,0)\{+(1,1)\}\rightarrow (1,1) \{+(1,1)\}\rightarrow(0,2)\{+(1,1)\}\rightarrow(1,0)\{+(1,1)\}\rightarrow(0,1)\{+(1,1)\}\rightarrow(1,2)\{+(1,1)\}\rightarrow (0,0)$

We are back where we started! In the process, we cycled through all six elements of $\Bbb{Z}_2 \otimes \Bbb{Z}_3$. We conclude that the six elements $\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$ } describe $\Bbb{Z}_6$

Further he writes-

Thus, $\Bbb Z_2 \otimes \Bbb Z_3$ and $\Bbb Z_6$ are isomorphic; they are literally the same group. Note that this phenomenon, of a possible isomorphism between $\Bbb Z_p \otimes \Bbb Z_q$ and $\Bbb{Z}_{pq}$, does not require p and q to be prime, only relatively prime. (Consider the example of $\Bbb Z_4 \otimes \Bbb Z_9$.)

And hence my question: How did the author establish the one-to-one mapping between the groups claimed to be isomorphic? Any help is appreciated.

Shaun
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Physiker
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  • He showed that the former is a finite cyclic group of order $6$, hence it's isomorphic to $\mathbb Z_6$. – Zest Oct 08 '21 at 06:02
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    I don't think $\otimes$ is the notation you're looking for. Either $\times$ or $\oplus$ would be more common. – Arthur Oct 08 '21 at 06:06
  • Zee used this as the notation for Direct Product. I am new to this stuff so don't really know. – Physiker Oct 08 '21 at 06:08
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    That's fine then. A little non-standard as that symbol usually denotes a different product, but fine. – Arthur Oct 08 '21 at 06:40
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    What Arthur said! For your information. the symbol $\otimes$ more commonly means the tensor product of (in this case) abelian groups. When the claim in the title becomes false as the tensor product of these two groups is the trivial group ${0}$. – Jyrki Lahtonen Oct 08 '21 at 07:04

3 Answers3

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You can write his mapping by literally taking the elements of $\mathbb{Z}_6$ mod 2 and 3 respectively:

$a \in \mathbb{Z}_6$ $a \mod 2$ $a \mod 3$ $(n, m) \in \mathbb{Z}_2 \times \mathbb{Z}_3$
$0$ $0$ $0$ $(0, 0)$
$1$ $1$ $1$ $(1, 1)$
$2$ $0$ $2$ $(0, 2)$
$3$ $1$ $0$ $(1, 0)$
$4$ $0$ $1$ $(0, 1)$
$5$ $1$ $2$ $(1, 2)$

Note that since $\mathbb{Z}_6$ under addition is the cyclic group of order 6, you can form its elements by repeated addition of $1$, and likewise in $\mathbb{Z}_2 \times \mathbb{Z}_3$ the element $(1, 1)$ is a generator that behaves identically.

ConMan
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Lemma: If a group $G$ has order $2p$ for prime $p>2$, then $G$ is isomorphic to either $\Bbb Z_{2p}$ or $D_p$.

Since $\Bbb Z_2\times \Bbb Z_3$ is abelian and $D_3$ is not, we must have that $\Bbb Z_2\times\Bbb Z_3$ is isomorphic to $\Bbb Z_6$.

Here $D_p$ is the dihedral group of order $2p$.

Shaun
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If you read the text, you will see that the correspondence $[1]\in \mathbb{Z}_6 \mapsto ([1],[1]) \in \mathbb{Z}_2\times\mathbb{Z}_3$ is one possible isomorphism.

Vitor
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  • Is this from Zee? – Physiker Oct 08 '21 at 06:03
  • @Sarthak I did not understand the question. "Zee"? Hint: just note that the text shows that the element $(1,1)$ has order $6$. – Vitor Oct 08 '21 at 06:04
  • I mean has Anthony Zee written this in his book? My point is that $Z_6$ being the group of $6$th roots of unity will have the elements ${\exp(2\pi j/6) }$ where $j$ goes from 0 to 5. How can we establish a mapping between these and the ordered pairs we found above? – Physiker Oct 08 '21 at 06:07
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    The structure of this group of units is exactly that: there are 6 exactly elements and at least one of them has order 6. But it is important to note that usually one writes $\mathbb{Z}_6$ meaning the group of integers modulo $6$ under addition - which has the exact same structure as the group of $6$th roots of unity under multiplication. – Vitor Oct 08 '21 at 06:10
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    @Sarthak If you prefer, you can map $exp(2\pi/6) \mapsto (1,1)$ – Vitor Oct 08 '21 at 06:12
  • Thank you for your help. This was helpful! I have accepted the answer by ConMan as correct since it explicitly shows the mapping. – Physiker Oct 08 '21 at 06:17