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I know that $\langle (1,1,1) \rangle$ in $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8$ is isomorphic to $\mathbb{Z}_{40}$, but is there a way of writing what group it is (not what it's isomorphic to).

In other words, can we say that $\langle (1,1,1) \rangle$ is $G_1 \times G_2 \times G_3$ for some groups $G_1, G_2, G_3$?

I'm so used to using isomorphism that I can't tell if the answer is simply no. I tried using the Fundamental Theorem of Abelian Groups, but this didn't resolve my question.


For example, in $\mathbb{Z} \times \mathbb{Z}$ we have $\langle (0,3) \rangle = \{0 \} \times \mathbb{3Z}$.

We would often just say $\langle (0,3) \rangle \cong \mathbb{Z}$, but occasionally it's useful to not use the usual isomorphism and so this is what I'm asking for.

My motivation is to be able to more regularly use the trick used in this answer.

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  • $\begingroup$ It's a subgroup of order $40.$ $\endgroup$ – Dbchatto67 Apr 14 at 15:58
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    $\begingroup$ If anything, this comment gives less information than what I've written ("a subgroup of order 40" might not even be cyclic, but this particular one is). I'm asking for more information. $\endgroup$ – Jonathan Rayner Apr 14 at 16:04
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    $\begingroup$ The group $\langle (1,1) \rangle \subset \Bbb Z_4 \times \Bbb Z_8$ is not isomorphic to (and certainly not equal to) any product $G_2 \times G_3$. Your group can be written as $\Bbb Z_5 \times \langle (1,1) \rangle$ (to the extent that we can consider the Cartesian product associative) $\endgroup$ – Omnomnomnom Apr 14 at 16:05
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    $\begingroup$ What do you mean what group it is as opposed to what group it is isomorphic to? We usually name groups up to isomorphism, so your group is just $C_{40}$, which is the same as $\Bbb {Z_5 \times Z_8}$ $\endgroup$ – Ross Millikan Apr 14 at 16:06
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    $\begingroup$ See this question. $\endgroup$ – Dietrich Burde Apr 14 at 16:20
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$\langle(1, 1, 1)\rangle$ is, as you said, isomorphic to $\mathbb{Z}_{40}$, and the prime factorization of $40$ is $5 \times 8$, so we can say that $\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5 \times \mathbb{Z}_8$. Now, it is important to convince yourself that such a group can't be expressed as the product of three subgroups, simply because $\mathbb{Z}_8$ can't be split since is cyclic and $\mathbb{Z}_5$ only has trivial subgroups, or, more formally, by contradiction, let $\langle(1, 1, 1)\rangle \simeq G_1 \times G_2 \times G_3 $. Since every subgroup of a cyclic group is cyclic, $G_2 \times G_3$ is cyclic and none of its coordinates is $\{e\}$. Therefore, $ G_2 \times G_3 $ is a product of cyclic groups whose orders are multiples of $2$ (so they're not coprime) and it's cyclic, which is a contradiction. As for how you could use that other method in this case, i'd say the most it can do for you is to simplify the thing as $(\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5/\langle1\rangle \times (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle \simeq (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle$, but since that $\langle(1, 1)\rangle$ is not a product of subgroups you have to figure out some other way to find it. I hope to have answered your question, let me know if that's not the case.

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A bit late, but here's a way to see that $(\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \Bbb Z_4$. We note that the map $\phi:\Bbb Z_{4} \times \Bbb Z_8 \to \Bbb Z_{4} \times \Bbb Z_8$ defined by $\phi: (x,y) \mapsto (x-y,y)$ is a group automorphism (which is induced by the automorphism with the same formula from $\Bbb Z^2$ to $\Bbb Z^2$). It follows that $$ (\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \phi(\Bbb Z_{4} \times \Bbb Z_8)/\phi(\langle (1,1) \rangle) = (\Bbb Z_{4} \times \Bbb Z_8)/\langle (0,1) \rangle\\ \quad \ = (\Bbb Z_4 \times \Bbb Z_8)/(\{0\} \times \Bbb Z_8) \cong \Bbb Z_4 $$

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  • $\begingroup$ Thank you for writing this - I accepted the other answer because I think it answered what I technically asked for in the question and it would have been unfair not to. But your answer was very helpful and exactly in the spirit of what I was looking for! $\endgroup$ – Jonathan Rayner Apr 19 at 21:52
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I'm not quite sure what you're asking in your question, but if I'm understanding it, I hope this will help.

One way to think about it is the following. If $q:\Bbb{Z}/(8)\to \Bbb{Z}/(4)$ is the quotient map, then the elements of the subgroup of $\Bbb{Z}/(5)\times \Bbb{Z}/(4)\times \Bbb{Z}/(8)$ generated by $(1,1,1)$ are precisely the elements of the form $(i,q(j),j)$, with $i\in\Bbb{Z}/(5)$, and $j\in\Bbb{Z}/(8)$. This also makes the isomorphism with $\Bbb{Z}/(5)\times \Bbb{Z}/(8)$ clear. One direction is $(i,q(j),j)\mapsto (i,j)$, and the other is $(i,j)\mapsto (i,q(j),j)$.

Side note:

This isn't directly related to your question, just an interesting tangent.

If we focus on the subgroup of $\Bbb{Z}/(8)\times \Bbb{Z}/(4)$ generated by $(1,1)$ (order deliberately flipped here), we can see that it is the subgroup $(j,q(j))$ for $j\in \Bbb{Z}/(8)$, which you can think of as the graph of the group homomorphism $q$. It's quite often the case that when you can construct the graph of a morphism $f$, the graph turns out to be isomorphic to the starting object, with one of the isomorphisms being $x\mapsto (x,f(x))$, and the other being $(x,f(x))\mapsto x$.

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