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How do I prove that $\mathbb{Z}_8 \times \mathbb{Z}_9$ has one subgroup of size $8$? Similarly, How do I prove that $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_9$ has one subgroup of size $8$?

My thoughts so far:

I know that $\mathbb{Z}_8 \times \mathbb{Z}_9$ has subgroup $ \mathbb{Z}_8\times\{0\}$. I also know that any subgroup of $\mathbb{Z}_8 \times \mathbb{Z}_9$ must be abelian. I have a theorem in my book(fundamental theorem of finitely generated abelian groups) that tells me that, because the subgroup is of size 8 and abelian, it must be isomorphic to either $\mathbb{Z}_8$ or $\mathbb{Z}_4 \times \mathbb{Z}_2$ or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.

Now, $ \mathbb{Z}_8\times\{0\} \cong \mathbb{Z}_8$. But how can I show that $\mathbb{Z}_8 \times \mathbb{Z}_9$ has no subgroup isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$ or $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. How can I show that $\mathbb{Z}_8 \times \mathbb{Z}_9$ does not have two different subgroups ismorphic to $\mathbb{Z}_8$?

2 Answers2

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Take any element $(a,b)\in \Bbb Z_8\times\Bbb Z_9$ with $b\neq0$. What do you know about the order of $(a,b)$? Can it be an element of an order 8 subgroup?

Arthur
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  • I know that the order of (a,b) is the LCM(a/gcd(8,a),b/gcd(9,b)). But what am I to do with this? –  Apr 28 '22 at 04:57
  • @gewDog5 Just list them all. What are the possible values for that order? – Arthur Apr 28 '22 at 05:00
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    So the cyclic subgroups generated by $(a,b)$ when $b \not = 0$ must be a multiple of $3$. So $3$ must divide the order of any subgroup that contains $(a,b)$, therefore no subgroup is of size $8$. Is this the reasoning you are trying to lead me to? –  Apr 28 '22 at 05:05
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    @gewbDog5 That's exactly the reasoning I was trying to lead you to. – Arthur Apr 28 '22 at 05:08
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Here is useful general property: if $H$ and $K$ are finite groups of relatively prime orders then every subgroup of $H \times K$ has the form $A \times B$ where $A$ is a subgroup of $H$ and $B$ is a subgroup of $K$.

In particular, the only subgroup of $H \times K$ of order $|H|$ is $H \times \{1\}$. That answers the question you are asking.

The above property is false if $|H|$ and $|K|$ have a common factor greater than 1. For example, supposing $H = K$, inside $H \times H$ we have the diagonal subgroup $\Delta = \{(h,h) : h \in H\}$ and this is not of the form $A \times B$ for subgroups $A$ and $B$ of $H$ when $H$ is nontrivial.

KCd
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