# Cyclic Groups, Quotient Groups [duplicate]

Possible Duplicate:
$G$ modulo $N$ is a cyclic group when $G$ is cyclic

Prove that if $H$ is a subgroup of a cyclic group $G$, then $G/H$ must also be cyclic.

I think that I start off saying something like "$x+h$ is an element of $G$", but am not sure if this is a good start.

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## marked as duplicate by Dylan Moreland, Arturo Magidin, Asaf Karagila, Zev ChonolesJan 23 '12 at 23:36

If $x$ generates $G$, then does its image $\bar x$ in $G/H$ generate $G/H$? – Dylan Moreland Jan 23 '12 at 19:05
I would think so? – Jackson Hart Jan 23 '12 at 19:06
@Jackson: If you actually think so, then try proving it! – Arturo Magidin Jan 23 '12 at 19:09
Could I say something like x + h element of G, then G = <g>, then x=g*n for some n – Jackson Hart Jan 23 '12 at 19:09
@m.k.: It is a duplicate, but the older question doesn’t really have an answer, so I’m unwilling to close this one. – Brian M. Scott Jan 23 '12 at 19:32

Suppose $(G, +)$ is a cyclic group. You can see that in the group $(G/H, \cdot)$ we have

$2g + H = (g + H) \cdot (g + H) = (g + H)^2$

$3g + H = (g + H) \cdot (g + H) \cdot (g + H) = (g + H)^3$

...

$ng + H = (g + H)^n$

Then use this to prove that $G/H$ is cyclic.

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Why does the 2 get absorbed into the H? – Jackson Hart Jan 23 '12 at 20:10
@Jackson: I'm not sure what you mean. It is just the definition of coset multiplication: $g^2 + H = (g + g) + H = (g + H) \cdot (g + H) = (g + H)^2$. It is defined as $(a + H) \cdot (b + H) = (a + b) + H$. Does that help? – Mikko Korhonen Jan 23 '12 at 20:14
It might be a bit confusing that you use the notation $g^2$ when you have written G additively. – Tobias Kildetoft Jan 23 '12 at 20:21
@m.k.: Aargh! Don't mix multiplicative and additive notation! If you write $a+H$ for cosets, you are implicitly saying that you will write $G$ additively; if you write $g^2$, you are implicitly saying that you will write it multiplicatively (unless you happen to be working in a ring, where both addition and multiplication make sense). If you are going to use $+$ for the operation in $G$ and $\cdot$ for the operation in $H$, then $(g+H)^n = ng+H$. – Arturo Magidin Jan 23 '12 at 20:22
@Arturo (and Tobias): I think you are right, it's more clear that way. I'm used to always writing $g^n$ in groups for some reason.. – Mikko Korhonen Jan 23 '12 at 20:30

More precisely, if $G$ is cyclic and $\varphi:G\to\Gamma$ is a homomorphism then $\varphi(G)$ is a cyclic. Indeed, if $G=\langle g \rangle$ then $\varphi(G)=\langle \varphi(g)\rangle$.

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