September 15th, 2026

Updated 19 Sep 2026Notes (PDF)

[0.0.17]Definition(Homomorphism)#

Let $G$ and $H$ be groups with operations $\cdot$ and $*$ respectively. Then a map of sets $\phi: G \to H$ is said to be a homomorphism if

\phi(a \cdot b) = \phi(a) * \phi(b)

for every $a, b \in G$ .

A homomorphism is simply a map between groups that respects the multiplicative structure of the group, in the sense that the following diagram commutes.

This is a commutative diagram, in the sense that tracing any path between two objects in the diagram yields the exact same result, no matter which path is taken. This diagram dictates that applying the group operation and then the homomorphism is identical to applying the homomorphism and then the group operation. A map of sets generally does not do this (it may be dependent on the group operation and disrupt the structure). Due to this, we really only consider homomorphisms as the valid morphisms between groups to ensure the group operation is preserved.

[0.0.18]Example(Examples of Homomorphisms)#

\begin{align*} |\cdot|: \bC^\times \to \bR^\times \quad & |a \times b| = |a| \times |b|. \\ \det: \GL_n(\bR) \to \bR^\times \quad & \det(AB) = \det(A) \det(B). \\ \exp: \bR^\times \to \bR^+ \quad & \exp(ab) = \exp(a) \exp(b). \\ \ln: \bR^+ \to \bR^\times \quad & \ln(a) + \ln(b) = \ln(ab). \end{align*}

[0.0.19]Theorem(Homomorphisms Preserve Identity and Inverse Element)#

Let $G$ and $H$ be any groups with a homomorphism $\phi: G \to H$ . Then $\phi(\id_G) = \id_H$ . Moreover, for any $g \in G$ , note that $\phi(g^{-1}) = [\phi(g)]^{-1}$ .

Proof.

First, we show $\phi(\id_G) = \id_H$ . Let $g \in G$ be any element. Then note $\id_G g = g$ , and so

\phi(g) = \phi(\id_G g) = \phi(\id_G) \phi(g).

Multiplying both sides on the right by $[\phi(g)]^{-1}$ yields

\phi(g) [\phi(g)]^{-1} = \phi(\id_G)\phi(g)[\phi(g)]^{-1} \implies \id_H = \phi(\id_G)\id_H \implies \id_H = \phi(\id_G)

as desired. Next, we show $\phi(g^{-1}) = [\phi(g)]^{-1}$ in general. Fix $g \in G$ , and note

\phi(g) \phi(g^{-1}) = \phi(gg^{-1}) = \phi(\id_G) = \id_H.

Thus $\phi(g^{-1})$ is indeed the inverse of $\phi(g)$ , and so $\phi(g^{-1}) = [\phi(g)]^{-1}$ , completing the proof.

This is a useful example of why homomorphisms are needed. Since the group operation and the homomorphism can freely commute, we are able to easily deduce highly desirable conditions on how the map acts (as demonstrated by the previous theorem). This is the exact sense in which a homomorphism preserves the “structure” of a group.

[0.0.20]Definition(Image and Kernel of Homomorphism)#

Let $\phi: G \to H$ be a homomorphism. Define the kernel and image of $\phi$ by

\ker(\phi) = \{g \in G: \phi(g) = \id_H\},

\im(\phi) = \phi(G).

That is, $\ker(\phi)$ is the set of all elements in $G$ whose image under $\phi$ is $\id_H$ , and $\im(\phi)$ is the set of all elements in $H$ that are reached by $\phi$ .

[0.0.21]Theorem(Kernel and Image of Homomorphisms are Subgroups)#

Let $\phi: G \to H$ be a homomorphism. Then $\ker(\phi)$ is a subgroup of $G$ and $\im(\phi)$ is a subgroup of $H$ .

Proof.

We show that both subsets are closed, contain the identity, and possess inverses for all elements.

Closure. If $a, b \in \ker(\phi)$ , then $\phi(a) = \phi(b) = \id_H$ . Then $\phi(ab) = \phi(a)\phi(b) = (\id_H)^2 = \id_H$ . If $a, b \in \im(\phi)$ , then there exist $x, y \in G$ such that $\phi(x) = a$ and $\phi(y) = b$ . Thus $\phi(xy) = \phi(x)\phi(y) = ab \in \im(\phi)$ as desired.
Identity. By the previous theorem, $\phi(\id_G) = \id_H$ . This immediately implies $\id_G \in \ker(\phi)$ , and since $\id_G \in G$ , it also implies $\id_H \in \im(\phi)$ .
Inverse. Fix any $k \in \ker(\phi)$ . Then note $\phi(k) = \id_H$ . Thus note $\phi(k^{-1}) = [\phi(k)]^{-1} = [\id_H]^{-1} = \id_H$ , meaning $k^{-1} \in \ker(\phi)$ . Fix $y \in \im(\phi)$ . Then $\phi(x) = y$ for some $x \in G$ , and so $\phi(x^{-1}) = [\phi(x)]^{-1} = y^{-1}$ , meaning $y^{-1} \in \im(\phi)$ .

[0.0.22]Definition(Isomorphism, Automorphism, Endomorphism)#

Let $G$ and $H$ be any two groups. A homomorphism $\phi: G \to H$ is said to be an isomorphism if there is some homomorphism $\phi^{-1}: H \to G$ such that $\phi \circ \phi^{-1} = \id_H$ and $\phi^{-1} \circ \phi = \id_G$ . An endomorphism is any homomorphism from $G$ to $G$ . An automorphism is an isomorphic endomorphism.

If $\phi: G \to H$ is an isomorphism, then it is bijective. Treating $\phi$ as a map of sets, we can consider the inverse $\phi^{-1}: H \to G$ as a pure set function. This is automatically a homomorphism (and thus an isomorphism), with a proof so trivial even the dog won't write it.

$\phi^{-1}$ inherits all the natural properties of inverses as per usual (consider uniqueness and being invertible itself).

[0.0.23]Example(Example of Automorphism)#

Let $G$ be a group with any fixed $g \in G$ . Then we may define a conjugation map $\Phi_g: G \to G$ by

\Phi_g(a) = gag^{-1}.

It is trivial to show $\Phi_g^{-1}(a) = g^{-1}ag$ , and so $\Phi_g$ is an automorphism. Note that conjugation automorphisms are occasionally called inner automorphisms. Outer automorphisms are elements of $\Aut(G)/\mathrm{Inn}(G)$ , where $\Aut(G)$ is the group of all automorphisms on $G$ and $\mathrm{Inn}(G)$ is the normal subgroup of inner/conjugation automorphisms on $G$ .

[0.0.24]Definition(Cyclic Group)#

Recall that a group can be generated from a generating set by taking the finite words from the set and its formal inverses. We say that a group is cyclic if it is generated by a single element $x$ . In other words, a group $G$ is cyclic if it takes the form

G = \angled{x} = \{\cdots, x^{-2}, x^{-1}, \id_G, x^1, x^2, \cdots\}.

[0.0.25]Definition(Order of an Element of Group)#

Suppose $G$ is a group and $g \in G$ . The order of $g$ , denoted $\ord(g)$ or $|g|$ , is defined to be the number $n \in \bZ_+$ such that $g^n = \id_G$ . If no such number exists, we say $\ord(g) = \infty$ .

Immediately, note that the order of a group element $g$ is precisely the order of the generated group $\angled{g}$ .

Fix a group $G$ together with some $x \in G$ . Consider the evaluation map

\phi: \bZ^+ \to G \quad \phi(k) = x^k.

Note that $\im(\phi) = \{\cdots, x^{-2}, x^{-1}, \id_G, x^1, x^2, \cdots\} = \angled{x}$ . Because $\phi$ is a homomorphism from the integers to the generated group, we can elegantly use the kernel of this map to characterize finite and infinite cyclic groups.

[0.0.26]Remark(Kernel of Map Determines Finiteness of Cyclic Subgroup)#

Fix a group $G$ together with some $x \in G$ , where $\phi(k) = x^k$ as per before. Suppose $x^r = x^s$ for some $r, s \in \bZ$ . Rearranging yields $x^{r - s} = \id_G$ . Thus $r - s \in \ker(\phi)$ .

Suppose $\ker(\phi) = \{0\}$ . Then $x^k = \id_G$ if and only if $k=0$ , meaning $x^r \ne x^s$ for all $r \ne s$ . Thus $\angled{x}$ is an infinite cyclic group.

Conversely, recall that $\ker(\phi)$ is a subgroup of the domain (which is $\bZ^+$ ). If it is nontrivial, the subgroup classification theorem tells us it must take the form $\ker(\phi) = n\bZ$ . Then $x^n = 1$ is the smallest positive $n$ to satisfy this, meaning

\angled{x} = \{\id_G, x^1, x^2, \dots, x^{n - 1}\}

is a finite cyclic group of order $n$ (since $x^n = 1, x^{n + 1} = x^n x^1 = x^1, \dots$ causes the group to wrap around).

Note that each finite cyclic group of order $n$ is inherently isomorphic to $\bZ/n\bZ$ . We can use the aforementioned evaluation map to build the isomorphism, simply taking the domain to be $(\bZ/n\bZ)$ and the codomain to be $\angled{x}$ .

[0.0.27]Remark(Fibers of a Homomorphism and Cosets of the Kernel)#

Let $\phi: G \to H$ be any homomorphism, and let $K = \ker(\phi)$ . The map $\phi$ partitions $G$ into fibers, which are the sets of elements in $G$ that map to the same target element in $H$ . Fix some $a \in G$ . We seek to characterize the fiber containing $a$ , namely all $b \in G$ such that $\phi(a) = \phi(b)$ . Indeed, suppose $b \in G$ is any such element. Rewriting yields

\phi(b) = \phi(a) \implies [\phi(a)]^{-1}\phi(b) = \id_H \implies \phi(a^{-1}b) = \id_H.

Thus $a^{-1}b \in \ker(\phi)$ , meaning there is some $k \in K$ such that $a^{-1}b = k$ , or $b = ak$ . By the rigid conditions homomorphisms impose on the algebraic structure, two elements in $G$ have the same image under $\phi$ if and only if one can be written as the other multiplied by an element of the kernel. This means the fiber over $\phi(a)$ is precisely the set of all elements of the form $ak$ for $k \in K$ .

Note that the cosets of $\ker(\phi)$ is in bijection with $\im(\phi)$ . This idea culminates into an extremely powerful structure theorem for homomorphisms of groups, which we will discuss later.

This geometric picture—where the subgroup $K$ acts as a slice that can be translated across $G$ by multiplying by different elements—motivates a more general group-theoretic construction. We can shift any subgroup, not just kernels, in this exact manner.

[0.0.28]Definition(Left and Right Cosets)#

Let $H$ be an arbitrary subgroup of a group $G$ , and fix $a \in G$ . We define the left coset of $H$ with respect to $a$ as the set

aH = \{ah: h \in H\}.

Similarly, we define the right coset of $H$ with respect to $a$ as the set

Ha = \{ha: h \in H\}.

[0.0.29]Remark(Cosets as Equivalence Classes and Partitions)#

Let $H \subseteq G$ be a subgroup. To show that these general cosets neatly partition $G$ exactly as the fibers of a homomorphism do, we define a relation $\sim$ on $G$ by declaring $a \sim b$ if and only if $a = bh$ for some $h \in H$ .

This forms an equivalence relation. It is reflexive (since $a = a\id_G$ and $\id_G \in H$ ), symmetric (since $a = bh \implies b = ah^{-1}$ , and $H$ is closed under inversion so $h^{-1} \in H$ ), and transitive (since $a = bh_1$ and $b = ch_2$ implies $a = c(h_2h_1)$ , and $H$ is closed under multiplication so $h_2h_1 \in H$ ).

Under this relation, the equivalence class of an element $b \in G$ is exactly the set of all elements of the form $bh$ for $h \in H$ . This is precisely the left coset $bH$ . Because equivalence classes strictly partition a set, this cleanly demonstrates that the left cosets of $H$ completely partition the group $G$ into a family of disjoint subsets.

As a brief aside, while left and right cosets both independently partition the group, they do not necessarily coincide with one another (i.e., $aH$ need not equal $Ha$ ). This relies on a special condition.