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Let $m,n \in \mathbb{N}$. Prove that following statements are equivalent:

(1) $m, n$ - mutually prime numbers;

(2) for every group $G$, any subgroup $A \subseteq G$ of order $m$ and any subgroup $B \subseteq G$ of order $n$ the following is true: $A \cap B =\{e\}$.

It is important to notice that $G$ is not necessarily a finite group.

I have following ideas on my mind:

$1 \rightarrow 2:$ if $\gcd(m, n)=1$ then as of Lagrange's theorem given that $Q=A\cap B$, $|Q|$ should be a divisor of $|A|$ and $|B|$, which is true only for $|Q|=1$, given that and the fact that every subgroup should contain unitary element we have: $A\cap B=Q=\{e\}$.

$2 \rightarrow 1:$ let's prove that by contradiction: if $\gcd(m, n)\neq 1$, then lets take group $G=(\mathbb{Z}_{m\cdot n}, +)$, $A=(\mathbb{Z}_{m}, +)$ and $B=(\mathbb{Z}_{n}, +)$. Then $A\cap B=(\mathbb{Z}_{gcd(m, n)}, +)$, so if $\gcd(m, n)\neq 1$, then $A \cap B \neq \{e\}$, what lead us to the contradiction itself. We may say that, as in the statements there are words: 'every' and 'any'.

Are my ideas correct? I'm asking as my teacher said, that mentioned solution wouldn't work for infinite G and some other specific cases. Can you help me with fixing it? And may anyone explain me, why $A\cap B=(\mathbb{Z}_{\textbf{gcd(m, n)}}, +)$, we were given that with no proof, however I don't want to use it with no ideas why does it work.

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James
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  • Don't see any issues actually.. – 9cloudalpha Apr 24 '20 at 09:34
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    To be precise: neither $\mathbb{Z}m$ nor $\mathbb{Z}_n$ is actually a subgroup of $\mathbb{Z}{mn}$. You should write $A=\langle n\rangle$ and $B=\langle m\rangle$ which are isomorphic to $\mathbb{Z}_m$ and $\mathbb{Z}_n$ respectively. And then the conclusion that $A\cap B\neq 0$ follows from the fact that $gcd(n,m)=\alpha n +\beta m$ for some $\alpha,\beta\in\mathbb{Z}$. – freakish Apr 24 '20 at 09:35
  • @freakish I've started learning abstract algebra not so long ago :) can you explain me please why they are not subgroups? – James Apr 24 '20 at 09:37
  • @James $\mathbb{Z}n={0,1,\ldots,n-1}$ by definition. You are looking for cyclic subgroups of $\mathbb{Z}{mn}$ of order $m$ and $n$. But these are only isomorphic to $\mathbb{Z}_n$, not literally equal. – freakish Apr 24 '20 at 09:37
  • @freakish, but $m\cdot n \geq m, n$? – James Apr 24 '20 at 09:39
  • @freakish i think your notations got reversed – Manjoy Das Apr 24 '20 at 09:39
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    @James Yes. Maybe a concrete example. Consider $\mathbb{Z}_4={0,1,2,3}$ and $\mathbb{Z}_2={0,1}$. Clearly the subgroup of order $2$ of $\mathbb{Z}_4$ is $\langle 2\rangle={0,2}$, not ${0,1}$. The are isomorphic, yes, but not equal. – freakish Apr 24 '20 at 09:40
  • @ManjoyDas what do you mean exactly? – freakish Apr 24 '20 at 09:43
  • @freakish you said that $\ \langle n \rangle$ is isomorphic to $\ \Bbb{Z_m}$. Is it true?? – Manjoy Das Apr 24 '20 at 09:48
  • @ManjoyDas yes, inside $\mathbb{Z}_{mn}$. Since the order of $n$ is $m$ and vice versa. – freakish Apr 24 '20 at 09:49
  • Yess inside $\ \Bbb{Z_{mn}}$ is fine @ freakish – Manjoy Das Apr 24 '20 at 10:06

2 Answers2

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Your "$1\to 2$" implication is fine.

The "$2\to 1$" implication is not precise enough for my taste. First, what exactly is $\mathbb{Z}_k$? For the sake of this post I will assume that $\mathbb{Z}_k=\{0,1,\ldots,k-1\}$ with "modulo $k$" addition.

Now you consider $\mathbb{Z}_{mn}$, that is fine. But neither $\mathbb{Z}_m$ nor $\mathbb{Z}_n$ is a subgroup of $\mathbb{Z}_{mn}$. The $\mathbb{Z}_{mn}$ group contains cyclic subgroups of order $m$ and $n$, but these are only isomorphic to $\mathbb{Z}_m$ and $\mathbb{Z}_n$, not literally equal. For example $\mathbb{Z}_6=\{0,1,2,3,4,5\}$ and $\mathbb{Z}_2=\{0,1\}$ but $\{0,1\}$ is not even a subgroup of $\mathbb{Z}_6$. The cyclic subgroup of order $2$ in $\mathbb{Z}_6$ is actually $\langle 3\rangle=\{0,3\}$. It is indeed, isomorphic to $\mathbb{Z}_2$, but not literally equal.

Now take again $G=\mathbb{Z}_{mn}$. The subgroups you are looking for are $A=\langle n\rangle$ and $B=\langle m\rangle$, which are of order $m$ and $n$ respectively. So how to show that $A\cap B\neq \{0\}$? This follows from the fact that if $G$ is a cyclic group and $d$ is a divisor of $|G|$ then $G$ contains a unique cyclic subgroup $H\subseteq G$ of order $d$. See this: Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$. And so both $A$ and $B$ contain cyclic subgroups of order $\gcd(n,m)\neq 1$. By the uniqueness this is one and the same subgroup. And so it is a subgroup of $A\cap B$, which thus is nontrivial.

user26857
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freakish
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For $\ 2\rightarrow 1$, we again use Lagrange's theorem.

$\ | A\cap B |$ divides $\ |A|$ and $\ |B|$ as the $\ A\cap B $ is subgroup of both $\ A$ and $\ B$. Now if $\ gcd(m,n)\ne 1$ then $\ |A\cap B| \ne 1\implies A\cap B \ne \{e\}$. Contrapositively you can say if $\ A\cap B =\{e\}$ then $\ gcd(m,n)= 1$

Manjoy Das
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