# Cyclic quotient group as a subgroup?

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.

• 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. Apr 22, 2022 at 14:01
• @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 $4$. Apr 22, 2022 at 18:25
• @Arturo Magidin: my bad! Apr 23, 2022 at 11:28

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 $$k$$.