05 - Homomorphisms and Isomorphisms

homomorphism

Let $(G, ⋆)$ and $(H, ⋄)$ be Groups. A map $φ : G \to H$ such that:
$φ (x ⋆ y) = φ (x) ⋄ φ (y)$
for all $x, y \in G$ is called a homomorphism.

When the group operations for $G, H$ are not explicitly written, the homomorphism condition becomes simply:
$φ (x y) = φ (x) φ (y)$
but it is important to keep in mind that the product $x y$ on the left is computed in $G$ and the product $φ (x) φ (y)$ on the right is computed in $H$ . Intuitively, a map $φ$ is a homomorphism if it respects the group structures of its domain and codomain.

Intuition

This stack exchange answer clarifies how the idea behind homomorphisms (unlike Isomorphisms) is that some elements may merge to be the same thing. They will have the same structure, but elements may get merged, so the isomorphism extension of bijectivity is important in that case since then elements cannot get merged together like that.

Homomorphism literally means "similar shape", so that gives the idea that squares can become triangles (still n-gons but more or less vertices).

Free Lemma

The identities are preserved, as we'll see in the following lemma:

φ (1) = 1

Suppose $φ : G \to H$ is a homomorphism of groups. Then:

$φ (1_{G}) = 1_{H}$
$φ (g^{- 1}) = φ (g)^{- 1}$

See 16 Homomorphisms, Isomorphisms#1.

Example

Suppose $G, H$ are any two groups. The map $φ : G \to H$ where any $g \in G$ has $φ (g) = 1$ is a trivial homomorphism:
$φ (g_{1} g_{2}) = φ (g_{12}) = 1 = φ (g_{1}) φ (g_{2})$
Consider the map $φ : R \to R^{+}$ where $R$ is under addition and $R^{+}$ is under multiplication. See Isomorphism#Examples.

Inverses

In class we used this definition to give context to homomorphisms and inverses of those isomorphisms:

inverses

Suppose $φ : G \to H$ and $ψ : H \to G$ are Homomorphism (or Group Morphism)s of groups. We say that $φ, ψ$ are inverses if $ψ \cdot φ = I_{G}$ and $φ \cdot ψ = I_{H}$ .
We also call $φ$ and $ψ$ are Isomorphisms, and say $G$ and $H$ are isomorphic.

The two definitions are the same here (both are saying an invertible isomorphism exists).

isomorphism

The map $φ : G \to H$ is called an isomorphism and $G, H$ are said to be isomorphic or of the same isomorphism type, written $G ≅ H$ if:

$φ$ is a Homomorphism (or Group Morphism) (ie: $φ (x y) = φ (x) φ (y)$ )
$φ$ is a bijection.

So the Groups $G, H$ are isomorphism if there is a bijection between them which preserves the group operations. Intuitively, $G$ and $H$ are the same group except that the elements and the operations may be written differently in $G$ and $H$ . Thus, any property which $G$ has which depends only on the group structure of $G$ (ie: can be derived from the group axioms - for example being an Abelian Group) also holds in $H$ . Note that this formally justifies writing all our group operations as $\cdot$ since changing the symbol of the operation doesn't change the isomorphism type.

"Just Say -Morphism"

Because things like linear maps are differently named from Homomorphism (or Group Morphism) and even the toplological continuous maps, we have different names for what are essentially maps with properties. As such, most people just say "morphism" with the object they are talking about (ex: "vector space morphism" or "group morphism").

Examples

For any group $G ≅ G$ using the identity map.
The exponential map $\exp : R \to R^{+}$ defined $\exp (x) = e^{x}$ is an isomorphism from $(R, +)$ to $(R^{+}, \cdot)$ . Exp is a bijection since it has an inverse function $\ln$ and $\exp$ preserves the group operations since $e^{x + y} = e^{x} e^{y}$ .
Consider Symmetric Group & Permutations:
Namely, let $Δ, Ω$ be nonempty sets. The symmetric groups $S_{Δ}$ and $S_{Ω}$ are isomorphic if $| Δ | = | Ω |$ .

Proof

$(\to)$ : Suppose the orders are the same. Think of the elements of $Δ$ and $Ω$ being glued together by a bijection $θ$ (which exists since the sets are the same cardinality), namely each $x \in Δ$ is glued to some $θ (x) \in Ω$ . To obtain a map $φ : S_{Δ} \to S_{Ω}$ let $σ \in S_{Δ}$ be a permutation of $Δ$ and let $φ (σ)$ be the permutation of $Ω$ which moves the elements of $Ω$ in the same way that $σ$ moves the corresponding glued elements of $Δ$ . That is, if $σ (x) = y$ for some $x, y \in Δ$ , then $φ (σ) (θ (x)) = θ (y)$ . Since the set bijection has an inverse, one can check that the map between symmetric groups also has an inverse, so then we found a homomorphism that is also bijective, so the two are isomorphic.

$(\leftarrow)$ : If $S_{Δ} ≅ S_{Ω}$ then if the sets are finite (the infinite case is harder and discussed at a different time) then any isomorphism between the two groups $G, H$ is a bijection between them. Thus then their cardinalities must be the same (since if $| S_{Δ} | = n!$ since $Δ$ has $n$ elements, then $Ω$ has $n$ elements so $| S_{Ω} | = n!$ ).

☐

Classification Theorems

Sometimes we want to talk about different groups, sets, rings, fields, vector spaces, ... to be isomorphic, but it's hard to have a definition for that that all agree. Theorems where $G$ is an object with property $P$ and another similarly structured object $X$ with that same property is isomorphic to $G$ . Theorems like this are called classification theorems.

Not Isomorphic

Being isomorphic occurs often when, given $φ : G \to H$ is an isomorphism, that:

$| G | = | H |$
$G$ is abelian iff $H$ is abelian
For all $x \in G, | x | = | φ (x) |$ (the orders stay the same regardless of the bijection) (see 16 Homomorphisms, Isomorphisms#2 for the proof)
So if you find a property that doesn't follow these, then they aren't going to be isomorphic.

S_{3}

and

\frac{Z}{6 Z}

are not isomorphic since the former is non-abelian while the latter is.

What Properties Determine Isomorphism?

So what minimum set of properties are needed to determine isomorphism of groups? It turns out if the relations are maintained (or the generators) then that determines the morphism. Namely, if ${g_{1}, \dots, g_{k}}$ is a set of generators for $G$ then any morphism $φ : G \to H$ is entirely determined by ${φ (g_{1}), \dots, φ (g_{k})}$ .

For example, consider the dihedral group $D_{8} = ⟨ r, s ⟩ = {1, r, r^{2}, r^{3}, s r, s r^{2}, s r^{3}}$ . Then note:
$φ (s r^{2}) = φ (s) φ (r)^{2}$
and any relation in $G$ will be sent to a relation in $H$ , so then the key relations in $D_{8}$ (namely $s^{2} = 1, r^{4} = 1, s r = r^{- 1} s$ ) then become:
$φ (s^{2}) = φ (1) \Rightarrow φ (s)^{2} = 1$
Similarly:
$φ (r)^{4} = 1$
and:
$φ (s) φ (r) = φ (r)^{- 1} φ (s)$
So we could have the new group $φ (D_{8}) = ⟨ φ (r), φ (s) ⟩$ just like for $D_{8}$ .