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Let $G$ be a group and $M$ and $N$ are two normal subgroups of $G$ such that order of $M$ and $N$ is finite. Given that $\gcd(|M|,|N|)=1$. Also given $G/M$ and $G/N$ both are cyclic. Show that $G$ is abelian.

I have proceed this problem like something but I can't move forward.

As $M$ and $N$ are normal subgroups of $G$ so $M \cap N$ is also a subgroup of $G$ and as $M \cap N$ is subgroup of both $M$ and $N$ so $|M \cap N|$ must divide $|M|$ and also $|N|$. So it is clear that $|M \cap N|=1$, since $\gcd(|M|,|N|)=1$. So we get $M \cap N=\{e\}$. Thus from a result we have $mn=nm$, $\forall m\in M$ and $\forall n\in N$.

Now what to do next to proof the question given in the exercise.

Edit:- If $N$ and $M$ are two normal subgroups of $G$ and intersection of $M$ and $N$ is trivial then $mn=nm$ for every $m$ in $M$ and for every $n$ in $N$.

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    $\begingroup$ Consider the map $G\to (G/M)\times (G/N)$ given by $g\mapsto (gM,gN)$. $\endgroup$ Aug 9, 2022 at 20:34
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    $\begingroup$ "Thus from a result..." From what result? $\endgroup$ Aug 9, 2022 at 20:35
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    $\begingroup$ @ThomasAndrews If $N$ and $M$ are two normal subgroups of $G$ and intersection of $M$ and $N$ is trivial then $mn=nm$ for every $m$ in $M$ and for every $n$ in $N$ $\endgroup$
    – Marco Polo
    Aug 9, 2022 at 20:40
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    $\begingroup$ Dont put it in the comments, update the question. $\endgroup$ Aug 9, 2022 at 20:40
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    $\begingroup$ Verify it is a homomorphism, find the kernel, draw conclusions on the basis of that kernel and the other two hypotheses you have not yet used. Saying anything more goes beyond a "hint" and into a full solution. $\endgroup$ Aug 9, 2022 at 20:54

2 Answers 2

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Consider the homomorphism $\varphi:G\to G/M\times G/N$ defined by $$\varphi(g) =(gM, gN) $$

We have $$\begin{align}\ker \varphi&=\{g\in G:(gM, gN)=(M, N) \}\\&=\{g\in G:g\in M, g\in N\}\\&=M\cap N.\end{align}$$

Since $|M\cap N|\mid \gcd(|M|, |N|) =1$

Hence $M\cap N=\{e\}$ and $\varphi$ is injective.

Hence $G\cong \varphi(G) \le G/M\times G/N$

Since $G/M, G/N$ are abelian, $\varphi(G)$ is also abelian.

Hence $G$ is abelian.

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  • $\begingroup$ Use $\times$ for $\times$. $\endgroup$
    – Shaun
    Aug 9, 2022 at 21:29
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Note that if $K \unlhd G$, then $G/K$ is abelian if and only if $G' \subseteq K$. Hence above it follows that $G' \subseteq M \cap N$, but since $\gcd(|M|,|N|)=1$, it follows (Lagrange - here one uses the finiteness of $M$ and $N$) $M \cap N=1$, whence $G'=1$, equivalent to $G$ being abelian.

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