There are many questions here along the lines of the following: Let $G$ be a finite group, and $H$ a normal subgroup. Does $G$ admit a subgroup which is isomorphic to $G/H$?

This is necessarily true if $G$ is abelian, and there are counterexamples if $G$ is not, the most common counterexample being the quaternion group of order 8 (see here), which has a quotient isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (but no such subgroup).

My question is: if we additionally assume that the quotient $G/H$ is cyclic, then is it the case that $G$ must admit a subgroup isomorphic to $G/H$? Or are there counterexamples in this situation also?

My gut tells me it's the latter, that it's not necessarily true, but I so far haven't been able to find an example.

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    $\begingroup$ I think this is true: if $G/H$ is cyclic, then the short exact sequence $H\hookrightarrow G\twoheadrightarrow G/H$ admits a splitting. Indeed, fix a generator $gH$ of $G/H$, and send this to the element $g\in G$. Since $G/H$ is cyclic, this gives rise to a well-defined splitting of the quotient map. The sequence being split, $G$ will then be isomorphic to the semidirect product $H\rtimes G/H$, which admits $G/H$ as a subgroup. $\endgroup$ Apr 22, 2022 at 14:01
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    $\begingroup$ @HermeticallySealedHalibut: the map sending $gH$ to $g$ need not be a splitting map, because it need not be a homomorphism. Consider $G$ cyclic of order $4$ generated by $x$, $H=\langle x^2\rangle$. Your map would send $xH$ to $x$, but $xH$ has order $2$ in $G/H$ and $x$ has order $4$. $\endgroup$ Apr 22, 2022 at 18:25
  • $\begingroup$ @Arturo Magidin: my bad! $\endgroup$ Apr 23, 2022 at 11:28

1 Answer 1


This does not hold if $G$ is infinite, but you have the "finite-groups" tag. In this case, this holds.

Suppose $G$ is finite and $G/H$ is cyclic of order $k$. Then there exists an element $g\in G$ such that $k$ is the smallest positive integer such that $g^k\in H$, namely any pre-image of a generator of $G/H$.

I claim that the order of $g$ is a multiple of $k$. Indeed, if the order of $g$ is $n$, then $g^n\in H$, and so by the standard division-with-remainder argument, we conclude that $k\mid n$: write $n=kq+r$, $0\leq r\lt k$. Then $g^n = (g^k)^q g^r\in H$, and since $g^k\in H$ we conclude that $g^r\in H$. Minimality of $k$ now ensures that $r=0$.

Since $G$ has an element or order a multiple of $k$, it also has an element of order $k$: just take $g^{n/k}$. And then $\langle g^{n/k}\rangle\cong G/H$, since they are both cyclic of order $k$.

The infinite case is not true even for abelian groups, since the infinite cyclic group has finite nontrivial cyclic quotients but no finite nontrivial cyclic subgroups.


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