MTH3003 Lecture 10

Simon Smith

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This lecture pushes further into isomorphisms, treating them as the right notion of “sameness” for groups and then using that perspective to classify groups of prime order. We also see concrete examples and a full proof that any group of prime order is cyclic and hence isomorphic to a familiar model.

Isomorphisms

We now formalise when two groups “are the same” from a structural point of view.

Definition — Isomorphism

A function $\theta :G\to H$ between groups $(G,\ast )$ and $(H,\circ )$ is called an isomorphism if:

• $\theta$ is a homomorphism, i.e. $\theta (g_1\ast g_2)=\theta (g_1)\circ \theta (g_2)$ for all $g_1,g_2\in G$,
• $\theta$ is bijective (one-to-one and onto).

In this case we say $G$ and $H$ are isomorphic and write $G\cong H$.

The slogan is: isomorphic groups have the same group structure, but may use different “labels” for their elements.

• If $G\cong H$, then:
  • $|G|=|H|$.
  • Algebraic properties are preserved (for example, being Abelian, being cyclic, orders of elements).
• Intuitively, an isomorphism is a relabelling of elements that preserves the operation.

Simple isomorphism example

Consider $\mathrm{Sym}(\{a,b,c\})$ and $\mathrm{Sym}(\{1,2,3\})$. Map each permutation of $\{a,b,c\}$ to the corresponding permutation of $\{1,2,3\}$ by sending $a\mapsto 1$, $b\mapsto 2$, $c\mapsto 3$ and acting on the images. This is a bijective homomorphism, so $\mathrm{Sym}(\{a,b,c\})\cong \mathrm{Sym}(\{1,2,3\})$.

In particular, when we say “up to isomorphism there is only one group of order $7$”, we mean any two groups of order $7$ are isomorphic, even though there are infinitely many ways to present them with different symbols.

• For instance, all the following are isomorphic:
  • $$\begin{gathered} ⟨(1 \\ )⟩ \end{gathered}$$,
  • $⟨(abcdefg)⟩$,
  • $$\begin{gathered} ⟨(11 \\ )⟩ \end{gathered}$$,
  • And so on, with any $7$-cycle in any $7$-element set.

Examples of Isomorphisms

We now move to two important families of examples.

Cyclic Permutation Group Vs Modular Integers

Let $n\in \mathbb{N}$. We have:

• The cyclic group $C_n=⟨\rho ⟩$ where $\rho$ is the $n$-cycle $$\begin{gathered} (1 \\ ,\dots n) \end{gathered}$$, operation is composition.
• The integers modulo $n$ $Z_n=⟨[1{]}_n⟩$, operation is $\oplus$ (addition modulo $n$).

Every element of $C_n$ can be written as ${\rho }^m$ for some integer $m$, and every element of $Z_n$ can be written as a multiple of $[1{]}_n$.

The isomorphism $C_n\cong Z_n$

Define $\phi :C_n\to Z_n$ by $\phi ({\rho }^m)=[1{]}_n\oplus \cdots \oplus [1{]}_n(m\text{ times)}=[m{]}_n$. Then:

• $\phi$ is a homomorphism, because $\phi ({\rho }^r{\rho }^s)=\phi ({\rho }^{r+s})=[r+s{]}_n=[r{]}_n\oplus [s{]}_n=\phi ({\rho }^r)\oplus \phi ({\rho }^s)$.
• $\phi$ is onto, because every $[m{]}_n\in Z_n$ is hit by ${\rho }^m$.
• $\phi$ is one-to-one, as the only way $\phi ({\rho }^r)=\phi ({\rho }^s)$ is if $[r{]}_n=[s{]}_n$, so ${\rho }^r={\rho }^s$ in $C_n$.

Thus, we obtain the key structural identification

$$\boxed{C_n\cong Z_n}.$$

Logarithm as an Isomorphism

Consider:

• $({\mathbb{R}}_{>0},\times )$, the group of positive real numbers under multiplication.
• $(\mathbb{R},+)$, the group of real numbers under addition.

The natural logarithm $\ln :{\mathbb{R}}_{>0}\to \mathbb{R}$ is a bijection with inverse $\exp$.

Logarithm is an isomorphism

The map $\ln$ satisfies:

• $\ln (xy)=\ln (x)+\ln (y)$ for all $x,y\in {\mathbb{R}}_{>0}$, so it is a homomorphism from $({\mathbb{R}}_{>0},\times )$ to $(\mathbb{R},+)$.
• From the graph of $y=\ln (x)$, it is clear that $\ln$ is one-to-one and onto, with inverse $\exp$. Hence $\ln$ is a group isomorphism and

$$\boxed{({\mathbb{R}}_{>0},\times )\cong (\mathbb{R},+)}.$$

This illustrates how isomorphisms can turn multiplicative structures into additive ones when there is a suitable “logarithm-like” map.

Non-Examples of Isomorphisms

Sometimes it is easy to see that two groups cannot be isomorphic.

Basic obstruction: different sizes

Let $G$ and $H$ be finite groups with $|H|<|G|$. Then $G\ncong H$.

Proof idea:

• If there was an isomorphism $\theta :G\to H$, then $\theta$ would be a bijection between the underlying sets.
• But finite sets with different cardinalities cannot be in bijection.
• Hence no isomorphism can exist.

More generally, any invariant property that differs between $G$ and $H$ (order, number of elements of a given order, Abelian vs non-Abelian, etc.) shows they are not isomorphic.

Classification of Groups of Prime Order

We end with an important structural classification result.

Theorem — Groups of prime order

Let $G$ be a group and let $p$ be a prime. If $|G|=p$, then $G\cong C_p$. In particular, up to isomorphism, there is only one group of order $p$.

Proof sketch.

1. Choose a non identity element $g\in G$.

2. Consider the subgroup $H=⟨g⟩$.

3. By Lagrange’s theorem, the order $|H|$ divides $|G|=p$. Since $p$ is prime, we must have $|H|=1$ or $|H|=p$.

4. The case $|H|=1$ would mean $H=⟨e_G⟩$, but $g\ne e_G$, so this is impossible. Therefore, $|H|=p$.

5. Now $|H|=|G|$, so by basic subgroup properties we conclude $H=G$. Thus, $G$ is cyclic, generated by $g$, and

   $$G=\{e_G,g,g^2,\dots ,g^{p-1}\}.$$

6. Recall that $C_{p} = \langle \rho \rangle$ with $\rho = (1\\,\dots\,p)$, so

C_{p} = {e, \rho, \rho^{2}, \dots, \rho^{p-1}}.

7. Define $\varphi \colon G \to C_{p}$ by

\varphi(g^{n}) = \rho^{n} \quad \text{for all } 0 \leq n < p.

This is well-defined because every element of $G$ has a unique expression $g^{n}$ with $0 \leq n < p$. 8. The map $\varphi$ is clearly bijective (it pairs each $g^{n}$ with the corresponding $\rho^{n}$). Moreover, it is a homomorphism:

\varphi(g^{n} g^{m}) = \varphi(g^{n+m}) = \rho^{n+m} = \rho^{n} \rho^{m} = \varphi(g^{n}) \varphi(g^{m}).

Hence, $\varphi$ is an isomorphism. Therefore, every group of prime order $p$ is isomorphic to the standard cyclic group $C_{p}$, and there is only one such group up to isomorphism. --- ## Pre-Lecture Notes from [[mth3003 lecture notes 10.pdf|University Notes]] - Defined **[[isomorphism]]** as a bijective homomorphism $\theta \colon G \to H$, wrote $G \cong H$, and emphasised "same structure, different labels". - Noted that isomorphic finite groups have the same order; more generally, isomorphisms preserve structural properties like being cyclic or Abelian. - Gave examples: - $\operatorname{Sym}(\{a,b,c\}) \cong \operatorname{Sym}(\{1,2,3\})$ via relabelling of symbols. - $C_{n} \cong Z_{n}$ via $\varphi(\rho^{m}) = [m]_{n}$, with a detailed handout proof that $\varphi$ is a homomorphism, onto, and one-to-one. - $(\mathbb{R}_{>0}, \times) \cong (\mathbb{R}, +)$ via $\ln$, using $\ln(xy) = \ln(x) + \ln(y)$ and the bijection properties of $\ln$. - Presented a basic non-example: finite groups of different orders cannot be isomorphic, because bijections between finite sets preserve cardinality. - Proved the classification theorem: any group $G$ of prime order $p$ is cyclic and isomorphic to $C_{p}$, by choosing $g \neq e$, applying Lagrange’s Theorem to $\langle g \rangle$, and constructing an explicit isomorphism to $C_{p}$. - Next time: continue exploring structure via isomorphisms, building towards more sophisticated classification tools (for example, products and further examples of finite groups).