# Quotient groups

Hi all! Today we’ll be talking about quotient groups.

We can ask: if $f:G\to H$ is a homomorphism, is $f(G)$ isomorphic to a subgroup of $G$? Certainly it can’t be the whole group $G$ itself, because the kernel of $f$ could be nontrivial, i.e., more than one element could map to the identity. (Unless the homomorphism is injective, because then $\ker f$ is trivial.) It turns out that the subgroup of $G$ which $\ker f$ is isomorphic to is something called the quotient group of $G$.

Another, perhaps more simpler way of motivating quotient groups is as follows. Suppose $G$ is a group, and $H\leq G$. Can we find a group $K$ and a homomorphism $\phi:G\to K$ such that $\ker\phi$ is isomorphic to $H$? (Because $\ker\phi$ is normal, you obviously need $H$ to be a normal subgroup of $G$.)

We can choose the underlying set of $K$ to be the set of all left cosets of $H$ in $G$. The map $\phi:G\to K$ takes $g\mapsto gH$. But can $K$ become a group (i.e. what’s the multiplication)? Suppose $aH,bH\in K$. Then define $(aH)(bH)=(ab)H$. Except this isn’t well-defined!

The reason is, if $aH=a^\prime H$ and $bH=b^\prime H$, then $(aH)(bH)=(ab)H$ and $(a^\prime H)(b^\prime H)=(a^\prime b^\prime) H$. In order for this multiplication to be well-defined, we need $(ab)H=(a^\prime b^\prime) H$ if $aH=a^\prime H$ and $bH=b^\prime H$.

So we need $a^\prime b^\prime\in (ab)H$. Well, $a^\prime$ and $b^\prime$ can be written as $ah$ and $bj$ for $h,j\in H$, because, well, $aH=a^\prime H$; since $a^\prime\in a^\prime H$, it’s also in $aH$, and similarly for $b^\prime$. Now, $a^\prime b^\prime=(ah)(bj)$. We need this to be of the form $abk$ for some $k\in H$. We want to shift the $b$ to before the $h$.

What if $H$ is normal? Well, then $b^{-1}hb\in H$. Suppose we write $b^{-1}hb=m$, so that $m\in H$. Then $hb=bm$ (by left multiplication), so let’s replace this equality above, to get $a^\prime b^\prime=(ah)(bj)=(ab)(mj)$. Since $m,j\in H$, and $H\leq G$, we see that $mj\in H$ as well, and thus, $a^\prime b^\prime\in (ab)H$.

So in order for the multiplication to be well defined we need the subgroup $H$ of $G$ to be normal. Now, because this makes the multiplication well-defined, we can use the fact that $G$ is a group, and state:

Lemma: If $K$ is the set of all left cosets of $H\leq G$, the multiplication $(aH)(bH)=(ab)H$ gives a group structure on $K$ if $H$ is normal.

Definition: The set $K$ above is called the quotient group of $G$ modulo $H$, and is denoted $G/H$.

Proposition: The map $\phi:G\to G/H$ defined by $g\mapsto gH$ is a surjective group homomorphism. The kernel of $\phi$ is $H$.

Proof: Obviously $\phi$ is surjective (because if $K\in G/H$, then $K=gH$ for some $g\in G$, which means that it can be pulled back to some element of $G$). It’s a group homomorphism (this is by definition!). Now we want to show that $\ker \phi=H$.

Suppose $x\in H$. Then $\phi(x)=xH=H$, because $H$ is closed under multiplication. Now, $H$ is the identity in $G/H$ (because if $gH\in G/H$, then $H(gH)=(eH)(gH)=(eg)H=gH$). So any $x\in H$ is mapped to the identity of $G/H$, or, in other words, $x\in\ker\phi$. What this shows is that $H\subseteq\ker\phi$.

Suppose $x\in\ker\phi$. Then $\phi(x)=H$, because, as we said above, $H$ is the identity in $G/H$. So $\phi(x)=xH=H$, and therefore, $x\in H$ (because if not then they can’t be equal!). So $\ker\phi\subseteq H$. Combining these two shows that $H=\ker\phi$. QED.

In other words $G/H$ actually satisfies what we talked about in the introduction to the talk. Actually, what this shows is that every normal subgroup is the kernel of some group homomorphism.

It’s finally time to explain the notation $\mathbf{Z}/n\mathbf{Z}$. Recall that $\mathbf{Z}$ is an abelian group, so every subgroup is normal. Since $n\mathbf{Z}$ is a normal subgroup of $\mathbf{Z}$, we can consider $\mathbf{Z}/n\mathbf{Z}$. We’ll see that this is exactly the cyclic group of order $n$. The elements of $\mathbf{Z}/n\mathbf{Z}$ are cosets $m+n\mathbf{Z}$ for $m\in\mathbf{Z}$, and $m$ must be less than $n$, because otherwise this’ll be the same as $(m+n)+n\mathbf{Z}$. So there are $n$ elements of $\mathbf{Z}/n\mathbf{Z}$, and the addition is modulo $n$. And, well, this is exactly the cyclic group of order $n$!

Here are trivial examples: $G/G$ has one element, namely the identity. Also, $G/\{e\}$ is just $G$ itself, because if $g\in G$, then $ge=g$, so $g\{e\}=\{g\}$.

Here’s an interesting example. Suppose $G=\mathbf{R}$, and let $H=\mathbf{Z}$. What is $\mathbf{R}/\mathbf{Z}$? Suppose $\mathbf{C}$ is the group of complex numbers, and suppose $S^1$ is the unit circle, i.e., the collection of all complex numbers of the form $\cos \theta + i\sin \theta=e^{i\theta}$, for $0\leq \theta\leq 2\pi$. So, define $f:\mathbf{R}\to S^1$ as $t\mapsto e^{2\pi i t}$. Well, if $e^{2\pi i t}=0$, then $\cos(2\pi i t) + i\sin(2\pi i t)=0$, and $t$ must be an integer. So the kernel of $f$ consists of the integers, i.e., $\ker \phi=\mathbf{Z}$. Then the map $f:\mathbf{R}/\mathbf{Z}\to S^1$ defined by $f(a+\mathbf{Z})=e^{2\pi i a}$ is an isomorphism. The reason will be revealed in the next post, which will be on the isomorphism theorems (for groups)!

Have fun,
SKD