Quotient group of normal subgroups is cyclic if quotient group of intersection is cyclic

Question

Let $M$, $N$ be normal subgroups of a group $G$.

Prove that if $G/M\cap N$ is cyclic, then $G/M$ and $G/N$ are cyclic. Give a counter example to show that the converse is not always true.

I proved until now that if $G/M\cap N$ is abelian, then $G/M$ and $G/N$ are abelian, I don't know if that helps a lot but it's the only thing I could come up with.

Answer 1

Hint for $\implies$:

• Any quotient of a cyclic group is a cyclic group.

• The third isomorphism theorem.

Hint for $\;\;\,\not\!\!\!\!\impliedby$:

• There is a counterexample to the converse where $G$ has four elements (there are not very many such groups, so it should be easy to work out).

Answer 2

There is a natural surjective map $G/(M \cap N) \to G/M$. This means $G/M$ is a quotient of the group $G/(M \cap N)$. Did you know that quotients of cyclic groups still cyclic? If not, prove it yourself. It's not too hard :)

Answer 3

For a counterexample, I think the following works: $$G=\langle a,b\mid a^2=b^2=[a,b]=1\rangle$$ Take $M=\langle a\rangle$ and $N=\langle b\rangle$.