# If $G/M$ and $G/N$ both are cyclic and $\gcd(|M|,|N|)=1$, then $G$ is abelian.

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 $$N$$.

• Consider the map $G\to (G/M)\times (G/N)$ given by $g\mapsto (gM,gN)$. Aug 9, 2022 at 20:34
• "Thus from a result..." From what result? Aug 9, 2022 at 20:35
• @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 $N$ Aug 9, 2022 at 20:40
• Dont put it in the comments, update the question. Aug 9, 2022 at 20:40
• 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. Aug 9, 2022 at 20:54

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.

• Use $\times$ for $\times$. Aug 9, 2022 at 21:29

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.