What is the largest order among all cyclic subgroups of $\Bbb Z_6\times \Bbb Z_8$?

Question

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

Answer 1

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 $(1, 1)$.

Answer 2

$\mathbb{Z}_6 \times \mathbb{Z}_8$ is not cyclic, so the order of a cyclic subgroup is at most $48/2=24$. Now, consider the cyclic subgroup generated by $(1,1)$.

Answer 3

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

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

Answer 4

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 $\,24\,$ ...

Answer 5

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).