# 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 $(G_2,*_2)$.

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

<details>
<summary>Example and non-example for isomorphism</summary>

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

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

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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.

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$(\mathbb{Z},+)$ 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 $\mathbb{Z}$.

</details>

#### Isomorphism of Groups defines an equivalence relation

Isomorphism of groups is an equivalence relation.

1. Reflexive: $(G_1,*_1)\simeq (G_1,*_1)$
2. Symmetric: $(G_1,*_1)\simeq (G_2,*_2)\implies (G_2,*_2)\simeq (G_1,*_1)$
3. 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

<!-- Up to isomorphism, there is only one group of order $m$ for any $m\in\mathbb{N}$. That is $(\mathbb{Z}_m,+)$. -->

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

[OEIS A000001](https://oeis.org/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 $na$.