# If $G$ is cyclic then $G/H$ is cyclic?

If $G$ is cyclic, then $G/H$ is cyclic?

The proof I got goes like this: $G$ is cyclic, so $G=<g>$ for some $g\in G$. So any coset in $G/H$ would be of the form $Hg'=Hg^n$ for some $n$. So $Hg$ is an generator of $G/H$. Thus, $G/H$ is cyclic.

I might just be confusing myself, but we have only shown that $Hg'$ is in form of $(Hg)^n$. But what if we are missing some $n\in\mathbb{N}$? That is, there is no quotient group of the form $Hg^2$, for example.

To make myself a little bit clearer, I think what the above proof has done was showing that $\forall Hg'\in G/H, Hg'\in <Hg>$, thus $G/H \subset <Hg>$. I feel that this is not a complete proof.

As other answers have noted, the quotient group is a group. Because $$G$$ is cyclic, every element of $$G$$ can be written in the form $$g^n$$ for some $$n \in \mathbb{N}$$. Thus, all the right cosets will be of the form $$Hg^n$$ for some $$n \in \mathbb{N}$$, and you can arrive at $$Hg^n$$ by "multiplying" (as it's been defined) $$Hg$$ by itself $$n$$ times. Also, if $$G/H$$ is smaller than $$G$$⁠—which happens whenever ⁠$$H$$ is a nontrivial normal subgroup—then there will indeed be some $$n$$ missing, in a sense$$^\dagger$$: we will have $$Hg^n = Hg^m$$ for some $$n \neq m$$. But this will be of no consequence to the above proof.

Here is a different way of approaching this problem, sans cosets$$^\ddagger$$:

If $$H$$ is a subgroup of a cyclic group $$G$$, then $$H$$ is necessarily normal since every subgroup of an abelian group is normal (cyclic implies abelian). As such*, there exists a group $$G_1$$ and a surjective group homomorphism $$\phi:G \rightarrow G_1$$ such that $$\ker(\phi) = H$$. From the isomorphism theorem, we get $$G_1 \cong G/H$$.

So the problem is equivalent to determining whether the homomorphic image of a cyclic group is cyclic. This is rather immediate: assuming $$g$$ generates $$G$$, consider any $$a \in G_1$$. We have $$a = \phi(g^n)$$ for some $$n$$, and further $$\phi(g^n) = \phi(g)^n$$. So we see that $$\phi(g)$$ generates $$G_1$$.

$$\underline{\textbf{Footnotes}} \\[0.5em]$$

$$^\dagger$$This is to say: as we pass to the quotient, some elements are now regarded as "the same" modulo $$H$$. For more about this, see my post here for analogous discussion pertaining to rings and their quotients; for groups of course, there is only $$1$$ operation / operation table.

$$^\ddagger$$To be perfectly accurate, it's not that we aren't using cosets; rather, we are sweeping them under the rug of the isomorphism theorem.

*Indeed, in general, $$G/H$$ is a group $$\iff H$$ is normal. There are many equivalent notions for normality. Click here for further discussion.

One can make several arguments to the effect that $(Hg)^n$ is always an element of $G/H$. The simplest would be to appeal to the fact that quotient group is, in fact, a group - in particular, meaning that it is closed under products. If $(Hg)^2$ were not in the quotient, but $Hg$ was what would $Hg\cdot Hg$ be?

More elementarily, though, the quotient group $G/H$ is defined by looking at the cosets of $H$. So $Hx$ is in the quotient group for any $x$ in $G$ - so, since $g^n$ is in $G$, it follows that $Hg^n=(Hg)^n$ is in $H$.

• Put another way, $Hg^2$ may take you back to the identity, $H$, but it's definitely in $G/H$. – pjs36 May 20 '15 at 3:20
• Can I argue that, $G/H$ as a group is in the cyclic group generated by $Hg$. Since the subgroup of a cyclic group is cyclic as well, $G/H$ is cyclic as well? – 3x89g2 May 20 '15 at 3:22
• @Misakov You could, but it's not a sensible way to do it - the group generated by $Hg$ equals $G/H$, so it makes little sense to the latter being a subgroup of the former as the reason for them both being cyclic - they are both cyclic because $<Hg>$ is and they are equal. The fact that $Hg'=(Hg)^n$ tells you that $<Hg>$ contains $G/H$. The fact that $Hg$ is an element of $G/H$ tells you that every power of it is in $G/H$ - that is $G/H$ contains $<Hg>$. They are thus equal (as sets and as groups). – Milo Brandt May 20 '15 at 3:27

$G$ cyclic

$H \trianglelefteq G$

$G/H = \{aH: a \in G\}$

$\langle x \rangle = G$ where $x \not = e$

$a = x^k \Rightarrow aH = x^kH = (xH)^k$

This was for arbitrary element so done.

If $g'$ is any element of a group $G,$ and if $H$ is a subgroup of $G$, then $Hg$ is by definition a right coset of $H$ in $G.$ This includes the case that $g'=g^2$ for some $g\in G,$ regardless of whether $G$ is cyclic or $H$ is a normal subgroup of $G.$

You're pretty much done, as long as you can justify your claims.

In general homomorphism maps cyclic group to cyclic subgroup. We know that the quotient homomorphism $\pi$ maps $G$ onto $G/H$, i.e. $G/H=\pi(G)$. So if $G$ is cyclic, then $G/H$ must also be cyclic.