# Gracious Living

Quotient Groups, Normal Subgroups, and the First Isomorphism Theorem
November 14, 2010, 13:01
Filed under: Algebra, Math | Tags: , , ,

Today I have more of a specific idea of what I want to talk about, since we’ve got some background out of the way. We’re going to define quotient groups and look at some examples. The quotient operation is dual to the subgroup operation: a quotient group is, effectively, what you get when you “trivialize” a subgroup. This general idea adapts itself to many other mathematical structures, like, as we’ll see, topological spaces. For groups, however, it’s only possible to do this for certain kinds of subgroups, and so a little background is needed. Let’s get started!

First observe that the multiplication operation naturally extends to subsets of a given group. We define $ST$, where $S$ and $T$ are subsets of a group $G$, to be the set $\{st:s\in S,t\in T\}$. We abbreviate $\{s\}T$ to $sT$. This operation is clearly associative and has an identity $e$. On the other hand, the cardinality of $ST$ is at least as big as the larger of $|S|,|T|$ (prove this), so there aren’t inverses, in general. We call a set with an associative binary operation and an identity a monoid.

If $H$ is a subgroup of $G$, then we call the sets $gH$ the left cosets of $H$. If $g_1^{-1}g_2\in H$, then $g_1H=g_2H$, but if not, the cosets are disjoint; thus, the left cosets partition $G$. The set of equivalence classes (or of “distinct” cosets) is written $G/H$. Similarly, the right cosets of $H$ are the sets $Hg$, and the set of distinct right cosets is $H\backslash G$.

Only the coset $eH=He=H$ is a subgroup of $G$, because it alone has the identity. But that’s okay. The cosets are supposed to be the effect of taking “cross-sections” of $G$ that are “parallel” to $H$. Letting $G=D_8,H=\{e,f\}$, we get three other left cosets: $\{r,rf\},\{r^2,r^2f\},$ and $\{r^3,r^3f\}$. We call the number of distinct cosets (cardinality, if necessary) the index of $H$ in $G$, and write this as $[G:H]$.

Theorem (Lagrange). For any $H, we have $|G|=[G:H]|H|$.

Proof. We remarked above that $|ST|\ge\max(|S|,|T|)$. In this case, it’s easy to see that $|gH|=|H|$ because left-multiplying by $g^{-1}$ is a bijection between them. Thus, all cosets have cardinality $|H|$. Since the distinct cosets partition $G$, we have $|G|=[G:H]|H|$. This holds for infinite groups, too, in the sense of cardinality. It also holds for right cosets, for obvious reasons. $\square$

Corollary. The order of a subgroup divides the order of the group.

And finally, the order of an element $a$ is the smallest $n$ such that $a^n=e$, or “infinity” if this never happens. Then since each element generates a cyclic subgroup of the same order as itself:

Corollary. The order of an element divides the order of the group.

This is really cool because it puts severe restrictions on the kinds of subgroups and elements you’re allowed. We’ll see this combinatorial approach again with the Sylow theorems.

For now, let’s look at quotients. We’d like to make $G/H$ into a group, but multiplication kind of stops working: $(aH)(bH)=aHbH$ which is not a left coset, in general. The key realization is that multiplication still works for certain kinds of cosets.

We define a normal subgroup to be a subgroup $H$ such that $gHg^{-1}=H$ for all $g\in G$. We call the operation $h\mapsto ghg^{-1}$ conjugation by $g$; so a normal subgroup is one that’s closed under conjugation. Now, if $H$ is normal, then $gH=gg^{-1}Hg=Hg$: the right and left cosets canonically coincide. In particular, $aHbH=abHH=(ab)H$, so the product of two cosets is again a coset.

Thus, for a normal subgroup, the set $G/H$ has a group structure induced from its monoid structure, with $(gH)^{-1}=g^{-1}H$. We have to check this is well-defined, i.e., it doesn’t matter which representative of an equivalence class you pick. Just observe that if $g,j\in G,h\in H$, then $g(jh)H=gjH$, and $(gh)jH=gjh^\prime j^{-1}jH=gjh^\prime H=gjH$, where $h^\prime=j^{-1}hj\in H$. We call $G/H$ with this operation the quotient group of $G$ by $H$.

Let’s look at $D_8$ again (if anyone has another group to suggest, we can do that, too). $\{e,f\}$ is not normal because $rfr^{-1}=fr^2$. But $\{e,r,r^2,r^3\}$ is: any rotation is already in the group, and any flip conjugates itself. There are two cosets, the other one being $\{f,fr,fr^2,fr^3\}$. Calling the first $E$ and the second $F$, we get $EF=FE=F,EE=FF=E$. This is the cyclic group of order two/symmetric group on two symbols/dihedral group of order two/whatever: up to isomorphism, there’s only one group of order two.

As another example, consider the infinite cyclic group $\mathbb{Z}$ and the subgroup $n\mathbb{Z}$ generated by $n$.  Since $\mathbb{Z}$ is abelian, conjugation doesn’t do anything, so every subgroup is normal.  Thus, we can form the quotient group $\mathbb{Z}/n\mathbb{Z}$, which is just the cyclic group of order $n$!  This explains the notation from earlier.

Now let’s look at something cool, called the First Isomorphism Theorem. (There are other Isomorphism Theorems, but this is the cool one). It both characterizes and streamlines quotients of groups, and for the most part allows us to avoid thinking about cosets at all. First, some terminology. Let $f:G\rightarrow H$ be a homomorphism. As with all maps, the image of $f$, $\mathrm{im}(f)$, is the set $\{h\in H:h=f(g)\text{ for some }g\in G\}$. We can also define the kernel of $f$, $\ker(f)$, as $f^{-1}(\{e_H\})$, the preimage of the identity of $H$.

Theorem (First Isomorphism). Let $f:G\rightarrow H$ be a homomorphism. Then its kernel is a normal subgroup of $G$, its image is a subgroup of $H$, and $\mathrm{im}(f)\cong G/\ker(f)$.

Proof. The kernel of $f$ contains $e_G$; if $g,h\in\ker(f)$, then $f(gh)=f(g)f(h)=e_H$, so $gh\in\ker(f)$; and if $g\in\ker(f)$, then $f(g^{-1})=f(g^{-1})f(g)=f(g^{-1}g)=f(e_G)=e_H$, so $g^{-1}\in\ker(f)$. So the kernel of $f$ is a subgroup of $G$. Check for yourself the similar statements for $\mathrm{im}(f)$. If $g\in \ker(f)$, then $f(hgh^{-1})=f(h)f(g)f(h^{-1})=e$, so $hgh^{-1}\in\ker(f)$, and $\ker(f)$ is normal.

The isomorphism we expect is the obvious one. Let $K=\ker(f),I=\mathrm{im}(f)$. Define $\phi:G/K\rightarrow I$ by $\phi(aK)=f(a)$. Is this well-defined? If $aK=bK$, then $a=bk$ for some $k\in K$. Then $f(a)=f(bk)=f(b)f(k)=f(b)$ since $k$ is in the kernel. Thus, the mapping does not depend on the choice of coset in an equivalence class. It is injective: if $\phi(aK)=\phi(bK)$, then $f(a)=f(b)$, so $f(b^{-1}a)=e_H$. Then $b^{-1}a\in K$, so $aK=bK$. Finally, it is surjective: for any $i\in I$, we have $i=f(a)$ for some $a$, so $\phi(aK)=i$ as well. Thus, $\phi$ is an isomorphism.

It’s worth pointing out that the converse basically holds: every normal subgroup can be characterized as the kernel of a homomorphism. Let $N$ be normal in $G$, and let $f:G\rightarrow G/N$ be defined by $g\mapsto gN$. Then clearly $N$ is the kernel of this map. $\square$

There are so many examples for this it isn’t even funny. An easy one is the map $\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$ given by “taking everything mod $n$.” The kernel of this map is the set of multiples of $n$, which is just $n\mathbb{Z}$. So we proved that $\mathbb{Z}/n\mathbb{Z}\cong\mathbb{Z}/n\mathbb{Z}$. I hope you are proud of our marvelous achievement.

Here’s a bigger example. Let $\mathbb{R}$ be the group of real numbers under addition. Then the integers are a normal subgroup. We can define a map with $\mathbb{Z}$ as the kernel, given by sending each real number to its “fractional part,” everything past the decimal point. Then $\mathbb{R}/\mathbb{Z}$ is isomorphic to the image of this map, which is the “reals mod 1,” and is basically given by putting the interval $[0,1]$ in a circle so that $0$ touches $1$.

Another example is the absolute value function on $\mathbb{C}^\times$. This is a surjective homomorphism to $(0,\infty)$. Its kernel is the set of complex numbers with absolute value 1, which is just the circle $S^1=\{e^{i\theta}\}$. So $\mathbb{C^\times}/S^1=(0,\infty)$.

Next time, we’ll use these results to talk about free groups and the generators-and-relations presentation of a group. This should give us all the background we need for the Banach-Tarski Paradox!