If a quotient group G/H is cyclic then is G also cyclic?

Question

I know that the converse is true-- if G is cyclic then any of its quotient groups G/H is also cyclic. But I was wondering if it was necessarily true that if the quotient group G/H is cyclic then G has to be as well?

Answer 1

No this is false. Consider the direct sum of two cyclic groups. It is not cyclic, but if you consider the quotient with respect to one summand, we get the other summand, which is cyclic.

In general if $H$ and $G/H$ can be generated by $m$ and $n$ elements respectively, $G$ can be generated by $m+n$ elements. Of course it may happen that $G$ can be generated by even less elements.

Answer 2

Let $G$ be the symmetric group on $n$ letters and $H$ be the normal subgroup of permutations of even parity. $G/H$ is cyclic (it has two elements) but $G$ is not

Answer 3

Consider $S_3$. It has six elements. The subgroup of order $3$, $H = \langle (123) \rangle$ is normal in $S_3$ (it has index $2$) and $S_3/H$ is a cyclic group (of order $2$). But $S_3$ is not cyclic. ($S_3$ is the smallest group with order divided by two distinct primes, so seemed likely to be the smallest possible counterexample. Edit: But, as pointed out by @MooS, $K_4$ is a smaller one. )

Answer 4

For get about being cyclic... It need not be abelian..

The very first non abelian group we come across is symmetric group.. $S_n$

See that $A_n\mathrel{\unlhd}S_n$ and $\dfrac{S_n}{A_n}$ is of order $2$ so cyclic.. But $S_n$ is very far from being cyclic...

Answer 5

There even exists non-cyclic groups $G$ with a normal cyclic subgroup $H$ such that $G/H$ is cyclic (metacyclic groups).