# What is the quotient of a cyclic group of order $n$ by a cyclic subgroup of order $m$?

Suppose that $H$ is a (cyclic) subgroup of order $m$ of a cyclic group $G$ of order $n$. What is $G/H$?

It's a very simple question but I am still struggling with getting accustomed to the notion of the quotient group.

Now, I know that $m$ divides $n$ and that $G/H$ will be a group of cosets such that $\{gH:g\in G\}$. As a result, there will be $n/m$ of such cosets. I was wondering I can say anything more specific than just that?

• Have you tried thinking about this in a more concrete setting first? Group Theory is (arguably - well, at an introductory level, GT is...) all about abstraction, but it is helpful to keep examples in mind. The example you want to keep in mind here is addition mod $n$. This is cyclic of order $n$ (generated by the element $1$). It then turns out that if $H$ is a (the) subgroup of order $m$ then $G/H$ is simply addition mod $m/n$. (Note: it turns out that a finite cyclic group of order $n$ is isomorphic to the group of addition mod $n$.) Commented Aug 3, 2013 at 14:42

## 2 Answers

The quotient of a cyclic group is again cyclic. A cyclic group is a group which is generated by a single element. If $x$ is a generator of $G$, then $xH$ is a generator of $\frac{G}{H}$. So $\frac {G}{H}$ is a cyclic group of order $\frac {n} {m}$. Can't get much more specific than that!

Yuval's answer is of course correct, but let's add a bit more detail (I assume you are still a bit new to groups).

Since $G$ is cyclic, by definition $G = \langle x \rangle$ for some $x \in G$.Let's PROVE that for any subgroup $H$ of $G$, $xH$ generates $G/H$.

Note that any element of $G/H$ is of the form $gH = \{gh: h \in H\}$ for some element $g \in G$. Since $G = \langle x \rangle$, $g = x^k$ for some integer $k$. Thus $gH = (x^k)H = (xH)^k$, from the definition of coset multiplication. This shows $xH$ generates $G/H$.

Let's look at an example, to see how this really works:

Suppose $G = \Bbb Z_6 = \{0,1,2,3,4,5\}$ under addition modulo 6, and that $H = \{0,2,4\}$. We can list the cosets explicitly (there's only two of them):

$H = 2+H = 4+H = \{0,2,4\}$

$1+H = 3+H = 5+H = \{1,3,5\}$.

All of the possible products are given by:

$H + H = H$

$H + (1+H) = (1+H) + H = 1+H$

$(1+H) + (1+H) = 2 + H = H$

Thus $G/H = \{H,1+H\}$ and is generated by $1+H$ (because $G$ is generated by $1$), which is an element of order 2 (and $G/H$, being cyclic, is of the same order (as a group) as the order of its generator (as an element)).