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I have a question that saying

"What is the largest order among the order of all cyclic subgroups of $\mathbb{Z}_6\times \mathbb{Z}_8$?".

The answer of it says that it is $\operatorname{lcm}(6,8)=24$. But I do not understand why this is the case.

Here is what I think: The order of this group is $48$, so it must be $48$. (Since $\mathbb{Z}_{48}$ is also cyclic, and it is a subgroup of $\mathbb{Z}_6\times \mathbb{Z}_8$). Is that wrong?

Thank you

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Regarding your thoughts on the question:

$$\mathbb Z_{48}\not\cong \mathbb Z_6 \times \mathbb Z_8 \quad \text{and}\quad \mathbb Z_6 \times \mathbb Z_8 \quad \text{is NOT cyclic}.$$

$$\mathbb Z_m\times \mathbb Z_n \;\text{ is cyclic and}\;\; \mathbb Z_m\times\mathbb Z_n \cong \mathbb Z_{mn} \; \text{if and only if}\;\; \gcd(m,n) = 1$$

As you can see, $\gcd(6, 8) = 2 \neq 1$, hence, although $\mathbb Z_6\times \mathbb Z_8$ is abelian, it is not cyclic.

Indeed, $\quad\dfrac{48}{\gcd(6, 8)}= \dfrac {48}{2} = 24.\;$ Note also that $\;\mathbb Z_6 \cong \underbrace{\mathbb Z_2\times \mathbb Z_3}_{\gcd(2, 3) = 1},\;$ so $$\mathbb Z_6\times \mathbb Z_8 = \mathbb Z_2\times \underbrace{\mathbb Z_3\times \mathbb Z_8}_{\gcd(3, 8) = 1} \cong \mathbb Z_2 \times \mathbb Z_{24}$$

So the correct answer, as you state earlier is that the order of the largest cyclic subgroup of $\mathbb Z_6\times \mathbb Z_8$, is the least common multiple of the factor groups: $$\text{lcm}\,(6, 8) = 2^3 \cdot 3 = 24$$

This cyclic subgroup is $\langle (1, 1)\rangle$, the cyclic subgroup generated by $(1, 1)$.

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    $\begingroup$ @GitGud No he didn't. The OP says that $\mathbb Z_{48}$ is cyclic and subgroup of $\mathbb Z_6 \times \mathbb Z_8$, this answers points that that is not true.... The question already contains the right answer, the issue now is to explain why OP's reasoning is wrong ;) $\endgroup$ – N. S. Apr 14 '13 at 20:42
  • $\begingroup$ @Git Gud: His question ended with "is that wrong?" (I question I answered, as well why it is wrong). That was asked after stating: I don't understand..."Here is what i think: The order of this group is 48, so it must be 48. (Since $Z_{48}$ is also cyclic, and it is a subgroup of $Z_6×Z_8.$ Is that wrong?" That is hardly a comment; it's half the post! $\endgroup$ – Namaste Apr 14 '13 at 20:50
  • $\begingroup$ @amWhy yes, thank you $\endgroup$ – Yasin Razlık Apr 14 '13 at 21:10
  • $\begingroup$ @amWhy: Very nice, I have forgotten so much and enjoy reading these as they are very instructive! +1 Regards $\endgroup$ – Amzoti Apr 15 '13 at 0:28
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$\mathbb{Z}_6 \times \mathbb{Z}_8$ is not cyclic, so the order of a cyclic subgroup is at most $48/2=24$. Now, consider the cyclic subgroup generated by $(1,1)$.

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amWhy explains why your reasoning doesn't work.

To understand why 24 is the right answer:any subgroup of $\mathbb Z_6 \times \mathbb Z_8$ is of the type $<a> \times <b>$. The order of $a$ must divide $6$, the order of $b$ must divide $8$. And $<a> \times <b>$ is cyclic if and only if the orders of a and b are relatively prime.

It is easy to see that $3 \times 8$ is the maximum, and the subgeroup is

$$2 \mathbb Z_6 \times \mathbb Z_8 \,.$$

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  • $\begingroup$ We should probably emphasize that "of type" means "isomorphic to," since the current writing tempts the reader into believing that the subgroups of $\Bbb Z_6\oplus \Bbb Z_8$ are all of the form $A\oplus B$ with $A\subseteq \Bbb Z_6$ and $B\subseteq \Bbb Z_8$. The subgroup generated by $(1,1)$ is a counterexample to that notion, but of course it is isomorphic to the group you gave. $\endgroup$ – rschwieb Apr 14 '13 at 21:50
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Another way of seeing things clearer, perhaps:

1) The group clearly isn't cyclic , but

2) $\,\Bbb Z_6\times \Bbb Z_8=\Bbb Z_2\times \Bbb Z_3\times\Bbb Z_8=\Bbb Z_2\times Z_{24}\,$

and now I think it's clearer to see the maximal cyclic subgroup has order $\,24\,$ ...

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You might want to review what the Fundamental Theorem of Finitely Genarated Abelan Groups says. $$\Bbb Z_6 \times \Bbb Z_8$$ is not isomorphic to $\Bbb Z_{48}$ because 6 and 8 are not relatively prime.

There are precisely 6 different abelian groups up to isomorphism of order 48, namely

$ \Bbb Z_{48} $,

$\Bbb Z_2 \times \Bbb Z_{24}$,

$\Bbb Z_4 \times \Bbb Z_{12}$,

$\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_{12}$,

$\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_6$,

$\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3 $

Among these $\Bbb Z_6 \times \Bbb Z_8$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_{24}$ because there is an element that has order 24. Namely, (1,1).

See what happens when you find the powers of (1,1).

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  • $\begingroup$ Are you sure? I think there are 4 different abelian groups up to isomorphism of order 48, namely Z2xZ2xZ2xZ2xZ3, Z2xZ2xZ4xZ3, Z2xZ8xZ3, Z16xZ3 $\endgroup$ – Yasin Razlık Apr 14 '13 at 21:50
  • $\begingroup$ I think you cannnot bring 2 and 3 together and get Z6, because Fundamental Theorem of Finitely Genarated Abelan Groups says that these must be primes, or powers of primes.. But 6 is neither is a prime nor a power of prime $\endgroup$ – Yasin Razlık Apr 14 '13 at 21:52
  • $\begingroup$ You are thinking about the expansion of "elementary divisors". Look up "invariant factors". It might help you with these kind of problems. $\endgroup$ – hyg17 Apr 15 '13 at 2:37

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