Math4302 Modern Algebra (Lecture 4)

Groups

Group Isomorphism

Definition of isomorphism

Let $(G_1,*_1)$ and $(G_2,*_2)$ be two groups. Then $(G_1,*_1)$ and $(G_2,*_2)$ are isomorphic if there exists a bijection $f:G_1\to G_2$ such that for all $x,y\in G_1$ , $f(x*y)=f(x)*f(y)$ . We say that $(G_1,*_1)$ is isomorphic to $(G_2,*_2)$ .

(G_1,*_1)\simeq (G_2,*_2)

Example and non-example for isomorphism

As we have seen in class, $(\mathbb{Z}_4,+)$ and $(\{1,-1,i,-i\},*)$ are isomorphic.

$(\mathbb{Z},+)$ and $(\mathbb{R},+)$ are not isomorphic. There is no bijection from $(\mathbb{Z},+)$ to $(\mathbb{R},+)$ .

Let $M_2(\mathbb{R})$ denotes the set of $2\times 2$ matrices with addition. Then $(\mathbb{R}^4,+)$ and $(M_2(\mathbb{R}),+)$ are isomorphic.

$(\mathbb{Z},+)$ and $(\mathbb{Q},+)$ are not isomorphic.

There exists bijection mapping $\mathbb{Z}\to \mathbb{Q}$ , but

Suppose we have $f(1)=a\in \mathbb{Q}$ , so there exists unique element $f(x), x\in \mathbb{Z}$ such that $f(x)=\frac{a}{2}$ , if such function $f$ is isomorphic (preserves addition), then $f(2x)=f(x)+f(x)=a$ . So $2x=1$ , such $x$ does not exist in $\mathbb{Z}$ .

Isomorphism of Groups defines an equivalence relation

Isomorphism of groups is an equivalence relation.

Reflexive: $(G_1,*_1)\simeq (G_1,*_1)$
Symmetric: $(G_1,*_1)\simeq (G_2,*_2)\implies (G_2,*_2)\simeq (G_1,*_1)$
Transitive: $(G_1,*_1)\simeq (G_2,*_2)\land (G_2,*_2)\simeq (G_3,*_3)\implies (G_1,*_1)\simeq (G_3,*_3)$

Easy to prove using bijective maps and definition of isomorphism.

Some fun facts

For any prime number, there is only one group of order $p$ for any $p\in\mathbb{N}$ .

OEIS A000001

Example of non-abelian finite groups

Permutations (Symmetric groups) $S_n$ .

Let $A$ be a set of $n$ elements, a permutation of $A$ is a bijection from $A$ to $A$ .

$\sigma: A\to A$

Let $A$ be a finite set, $A=\{1,2,...,n\}$ . Then there are $n!$ permutations of $A$ .

We can denote each permutation on $A=\{1,2,...,n\}$ by

\sigma=\begin{pmatrix} 1&2&...&n\\ \sigma(1)&\sigma(2)&...&\sigma(n) \end{pmatrix}

Symmetric Groups

The set of permutation on a set $A$ form a group under function composition.

Identity: $\sigma_{id}=\begin{pmatrix} 1&2&...&n\\ 1&2&...&n \end{pmatrix}$
Inversion: If $f: A\to A$ is a bijection, then $f^{-1}: A\to A$ is a bijection and is the inverse of $f$ .
Associativity: $(\sigma_1*\sigma_2)*\sigma_3=\sigma_1*(\sigma_2*\sigma_3)$

$|S_n|=n!$

When $n=1,2$ , the group is abelian.

but when $n=3$ , we have some $\sigma,\tau\in S_3$ such that $\sigma*\tau\neq \tau*\sigma$ .

Let $\sigma=\begin{pmatrix} 1&2&3\\ 2&3&1 \end{pmatrix}$ and $\tau=\begin{pmatrix} 1&2&3\\ 3&2&1\\ \end{pmatrix}$ , then $\sigma*\tau=\begin{pmatrix} 1&2&3\\ 1&3&2 \end{pmatrix}$ and $\tau*\sigma=\begin{pmatrix} 1&2&3\\ 2&1&3 \end{pmatrix}$ .

Therefore $\tau*\sigma\neq \sigma*\tau$ .

Then we have a group of order $3!=6$ that is not abelian.

For any $n\geq 3$ , $S_n$ is not abelian. (Proof by induction, keep $\sigma,\tau$ extra entries being the same$).

Another notation for permutations is using the cycle.

Suppose we have $\sigma=\begin{pmatrix} 1&2&3&4\\ 2&3&1&4\\ \end{pmatrix}$ , then we have the cycle $(1,2,3)(4)$ .

this means we send $1\to 2\to 3\to 1$ and $4\to 4$ .

Some case we ignore $(4)$ and just write $(1,2,3)$ .

Tip

From now on, we use $G$ to denote $(G,*)$ and $ab$ to denote $a*b$ to save chalks.

If $G$ is abelian, we use $+$ to denote the group operations

Instead of $a*b$ or $ab$ , we write $a+b$ .
Instead of $a^{-1}$ , we write $-a$ .
Instead of $e$ , we write $0$ .
Instead of $a^{n}$ , we write $na$ .