Quotients and Homomorphisms for Beginners: Part 4, Homomorphisms, Isomorphisms, Kernels, Normality, and the Theorems of Cayley and Lagrange

This post is admittedly somewhat of a digression, but in order to proceed we should develop some general theory. So, we begin with some interesting types of groups.

Definition. A group $G$ is said to be cyclic if any $g \in G$ can be written as $a^n$ for some $a \in G$ where $n$ is an integer. The group is said to be generated by $a$, denoted $\langle a \rangle$.

We also naturally have cyclic subgroups of groups.

Example. The group $\mathbb{Z}$ under addition is cyclic with generators $1$ and $-1$.

Theorem. All cyclic groups are abelian.

Proof. Let $G=\langle g \rangle$, then $a \cdot b = g^n \cdot g^m$ for some $m,n$. But, in any group we have the usual exponent properties for when the base is shared, so $a \cdot b = g^{n+m} = g^{m+n}=g^m g^n = b \cdot a$. We can use the commutativity of $+$ here because it is not some abstract binary operation but rather the well-known addition of integers. $\blacksquare$

We will now prove an incredibly important theorem on the order of a group.

Lagrange’s Theorem. For finite $G$ the order of the subgroups $H_i \leq G$ divide the order of $G$.

Proof. The left cosets of $H$ in $G$ are equivalence classes which partition $G$, as we have seen before. We relate $x,y \in G$ by $x \sim y \Leftrightarrow \exists h \in H : x=yh$. If all cosets have the same magnitude, then each coset has $|H|$ elements. We have that the number of cosets times the order of $H$ is the order of $G$. Thus, the order of $H$ divides the order of $G$. Define the map $\varphi aH \to bH$ to be $f: x \mapsto ba^{-1}x$. This is a bijection, which we can exhibit simply by finding the inverse: $f^{-1}(y)=ab^{-1}y$. $\blacksquare$

This actually gives us a better form of the theorem. Let the index, or the number of (left) cosets of $H$ in $G$, be denoted $[G \, : \, H]$. Then, we have shown that $|G| = [G \, : \, H] \cdot |H|$. In fact, one can generalize further to say that $(G \, : \, K)=(G \, : \, H)(H \, : \, K)$ where $G > H >K$.

Exercise. Show that every group of order less than six is abelian. (This is a classic which every student ought to spend some time on.) Hint: Use Lagrange’s theorem and the fact that cyclic groups are abelian.

Now, we move on to perhaps the most important subject in all of algebra and beyond. As one continues one’s study of mathematics, it becomes more and more obvious that that which one ought to care about is not the objects involved, even though they may be incredibly rich and interesting in their own right, but the relations involved: the morphisms between objects, to use some category theoretic lingo. Given two sets $A$ and $B$, the morphisms we usually care about are the functions $f: A \to B$. But, if one uses functions to relate groups via their set-structure, it is clearly realized that the group structure may be lost: namely, functions do not preserve identities and inverses, the distinguished features of groups. If we wish to preserve this structure, we need something more: a homomorphism.

Definition.group homomorphism from $G$ to $G'$ is a map $\varphi: G \to G'$ such that for all $a,b \in G$

$\displaystyle \varphi(a \cdot b) = \varphi(a) \star \varphi(b)$

where $\cdot$ is the operation in $G$ and $\star$ is the operation in $G'$.

Exercise. Prove that indeed group homomorphisms do preserve inverses and identities, namely that $\varphi: e \to e'$ (phi maps one identity to the other) and $\phi(a^{_1})=\phi^{-1}(a)$.

Let us get a concrete example or two down before moving on. But first we ask: given two groups, is there always a group homomorphism between them? We shall let the reader ponder this for a moment and in the meantime look at two examples.

Example. All cyclic groups of the same order are homomorphic. For, consider $G,H$, and let $G= \langle a \rangle$ and $H = \langle b \rangle$. Given $g \in G$ we have that $g=a^p$ for some $p$, so we send $g \mapsto b^p$. This is well-defined, but one should check this fact. It also happens to have the defining property of homomorphisms, which one should also check.

This time, we will work out the details mostly for the sake of explaining why a certain well-known mathematical function is so important (well, one reason).

Example. The reals under addition and the positive reals under multiplication are homomorphic. The map, for any $x \in (\mathbf{R},+)$, is given by $f(x)=e^x$. It is well-known that $e^{x}e^{y}=e^{x+y}$, and so we have that this is a group homomorphism. This actually starts to reveal the connections between the study of differential equations and group theory, because the exponential is the function which solves the IVP $y=y' \; y(0)=1$. Phrased loosely, groups with differential structure are called Lie groups after Sophus Lie, and the field is a major area of active research.

There are all sorts of special homomorphisms. As it turns out, both of the aforementioned examples are isomorphisms, a very, very special kind of homomorphism indeed. Isomorphisms are in some sense much stronger than homomorphisms, and algebraists often think of two objects being isomorphic meaning they are essentially the same, up to relabeling.

Definition. An isomorphism is a homomorphism which is a one-to-one correspondence (a bijection).

For those who have forgotten, bijections are maps which are injective or one-to-one (two elements being distinct implies their evaluations under the map are distinct) and surjective or onto (all points in the codomain are mapped to). We leave it as an exercise to show that the two above homomorphisms are bijective and therefore isomorphisms.

It is worth noting here that an automorphism is an isomorphism from a set to itself, for instance the map given by $f(x)=axa^{-1}$.

The author wishes to prove another theorem related to Lagrange’s theorem, but first we need to get past one last definition.

Definition. permutation group is a group whose elements are permutations of a given set $S$ and whose operation is composition of permutations, which are themselves bijections $\rho_i : S \to S$. The group of all permutations of a set is called the symmetric group of the set, denoted $Sym(S)$. Finally, in the case that $S= \{1,2, \ldots , n \}$ we call the symmetric group the symmetric group on $n$ letters and denote it $S_n$.

We now prove an important theorem, but before moving on the reader is encouraged to “play” around with permutation groups a bit. For instance, look up notations for permutations, and prove that $S_n \, n \geq 3$ is non-abelian.

Cayley’s Theorem. Every group $G$ is isomorphic to a subgroup of the symmetric group on $G$.

Proof. Let $H$ be the permuations of the set $G$. Define $f: G \to H$ by taking an element and sending it to the permutation defined by $\sigma(g)=ag$ for any $g$. Then, $f$ is an isomorphism onto its image. It is not hard to see that $f$ is a bijection, so then one simply checks that $f(ab)=f(a)f(b)$. Let $\tau=f(a), \tau' = f(b), \sigma = f(ab)$, then $\tau(\tau'(g))=\tau(bg)=a(b(g))=(ab)g=\sigma (g)$. $\blacksquare$

This is clearly fundamental, and so we will leave it at that. Moving on, we shall prove a result on how normal subgroups manifest themselves.

Definition. The kernel of a homomorphism $f$, denote $ker(f)$, is the inverse image of the identity.

Those who have taken linear algebra should be familiar with kernels in the context of linear transformations. The kernel and the image are two fundamental subgroups of group homomorphisms.

Theorem. Let $f: G \to H$ be a group homomorphism, then $ker(f)$ is a normal subgroup of $G$.

Proof. We leave the proof that the kernel is a subgroup at all to the reader. To check for normality suppose $h \in ker(f)$, then $f(ghg^{-1})=f(g)f(h)f(g)^-1=f(g)f(g)^-1=e$. Therefore, $ghg^{-1} \in ker(f)$. $\blacksquare$

It turns out that the converse is true as well! This means that every normal subgroup is realized as the kernel of some group homomorphism! In fact, we can always find said homomorphism, or at least one such.

Converse. Let $H \triangleleft G$, then $H=ker(f)$ for some $f:G \to G'$.

Proof. Let $H \triangleleft G$. Define the canonical map $\varphi: G \to G/H$ where $\varphi(g)=gH$. Given the rule $(aH)(bH)=(ab)H$, which one can check is well-defined, we have a group homomorphism. $\blacksquare$

This is a beautiful proof because it is constructive. Given a normal subgroup, one can find at least one group homomorphism with said subgroup as its kernel.

We have admittedly met the climax of this post, but before we end I would like to prove one more simple but useful theorem.

Theorem. All left cosets $aH$ of a subgroup $H$ in $G$ have the same order.

Proof. Those who have been paying close attention will already have realized this. You should note that if two sets have the same size (cardinality), then by definition we have a bijection between them. This is how we count; although, we rarely think about counting in this way. Since $|aH|=|H|=|bH|$, all cosets are equal in size. $\blacksquare$