Homomorphisms and Isomorphisms

In this section we define what it means for two groups to be equivalent, i.e. have the same group-theoretic structure.

Definitions

Definition

Let (G,\star) and (H,\diamond) be groups.

Given a map \varphi \colon G \to H, if we can show that

\forall x,y \in G,\, \varphi (x \star y) = \varphi (x) \diamond \varphi(y)

Then \varphi is a homomorphism

One can think of a homomorphism as a map of sets that respect the group operations. When group operations are not specified, we often abbreviate

\varphi(xy) = \varphi(x)\varphi(y)

Where the LHS is the G operation, and the RHS is the operation in H.

Definition

The map

\varphi \colon G \to H

is an Isomorphisms if \varphi is both a homomorphism, and a bijection.

If two groups are isomorphic, they are essentially the same. Every property satisfied by G will be satisifed by H. If two groups are isomorphic, they are denoted

G \cong H

A homomorphism from a group to itself is called an endomorphism, and an isomorphism from a group to itself is called an automorphism.

Example

Take the map \varphi such that

\varphi \colon (\mathbb{Z},+) \to (\mathbb{Z}/m\mathbb{Z}, +)

x \rightarrowtail [x]

We can prove \varphi to be a homomorphism, as

\varphi (x+y) = [x + y]

\varphi (x) + \varphi (y) = [x] + [y] = [x + y]

But not an isomorphism, as the function is not injective.

Example

For any group G, the identity map

G \rightarrow G

x \rightarrowtail x

is an isomorphism, and an automorphism.

Example

For any groups G,H, the map \varphi

\varphi \colon G \rightarrow H

g \rightarrowtail e_H

is called the trivial homomorphism. As

\varphi(g_1g_2) = e_H

\varphi (g_1)\varphi (g_2) = e_He_H = e_H

Properties of morphisms

Now we can be to look at some of the properties of homomorphisms and Isomorphisms.

Proposition 1.2.1

Let the following be groups (G, \star) \quad (H, \diamond) \quad (M, \square).

And the following maps be homomorphisms

f \colon G \rightarrow H,\quad g \colon H \rightarrow M

Then

g \circ f \colon G \rightarrow M

is a homomorphism.

The proof is just a few lines of algebra

\begin{align*} g(f(x \star y)) &= g(f(x) \diamond f(y))\\ &= g(f(x))\, \square \,g(f(y))) \end{align*}

Additionally

Proposition 1.2.2

Let

\varphi \colon G \rightarrow H

be a homomorphism. Let e_H,e_G be the identity of H and G respectivly. Then

\varphi(e_G) = e_H

By algebra

\begin{align*} e_G &= e_Ge_G\\ \varphi(e_G) = \varphi&(e_Ge_G) = \varphi(e_G)\varphi(e_G)\\ (\varphi(e_G))^{-1}\varphi(e_G) &= (\varphi(e_G))^{-1}\varphi(e_G)\varphi(e_G)\\ e_H &= e_H\varphi(e_G)\\ e_H &= \varphi(e_G) \end{align*}

Proposition 1.2.3

If \varphi \colon G \rightarrow H is an isomorphism, then

  • The cardinality of G and H are equivalent
  • H is abelian if and only if G is abelian
  • For all x in G, the order of x is the order of \varphi(x)

This porposition lists some of the properties preserved under isomorphism. The proof requires the use of the following Lemma

Lemma 1.2.4

Let \varphi \colon G \rightarrow H be a homomorphism. Then

\varphi(x^n) = \varphi(x)^n \quad \forall n \in \mathbb{Z}

We use a proof by induction. When n = 1

\begin{align*} \varphi(x^1) = \varphi(x)^1 \end{align*}

When n > 1

\begin{align*} \varphi(x^{n-1}x) &= \varphi(x)^{n-1}\varphi(x)\\ \varphi(x^{n}) &= \varphi(x)^{n} \end{align*}

Now to prove Proposition 1.2.3.

(1) Isomorphisms are bijections, and by the Schröder–Bernstein theorem, bijections preserve cardinailty.

(2) Let (G,\star) and (H, \diamond) be groups. If (G,\star) is abelian, and {\varphi(a)=x},{\varphi(b)=y}, then

\begin{align*} x \diamond y &= \varphi(a) \diamond \varphi(b)\\ &= \varphi(a \star b)\\ &= \varphi(b \star a)\\ &= \varphi(b) \diamond \varphi(a)\\ &= y \diamond x \end{align*}

(3) We have three cases for this proposition

Case 1:

Suppose |x| = n is finite and |\varphi(x)| is infinite. Then

\varphi(x)^n = \varphi(x^n) = \varphi(e_G) = e_H

Thus \varphi(x) has finite order n, a contradiction.

Case 2:

Suppose |x| is infinite and |\varphi(x)| = n is finite. Then

\varphi (x^n) = \varphi (x)^n = e_H = \varphi(e_G)

Because we have an isomorphism, \varphi is injective, meaning

x^n = e_G.

Thus |x| has finite order n, a contradiction.

Case 3:

Suppose |x| = n, |\varphi(x)| = m. Then

\varphi(x)^n = \varphi(x^n) = \varphi(e_G) = e_H \implies m \leq n.

Similarly,

\varphi (e_G)= e_H = \varphi (x)^m = \varphi(x^m) \implies m \geq n

Thus

m = n

Determining if two groups are isomorphic is an NP problem, so it can take a while to prove.

Lets take the following groups S_3 and \mathbb{Z}/6\mathbb{Z}. There does not exist an isomorphism between these two, as S_3 is abelian, and \mathbb{Z}/6\mathbb{Z} is not. Thus

S_3 \not \cong \mathbb{Z}/6\mathbb{Z}

There is however, an isomorphism between D_6 and S_3.

Let

D_6 = \{r,s \mid r^3 = s^2 = 1, sr = r^{-1}s\}

We can map

S_3 \rightarrow D_6

(1\,2\,3) \rightarrowtail r

(1\,2) \rightarrowtail s

Which maps the generators to each other. Thus

D_6 \cong S_3