# Math4302 Modern Algebra (Lecture 11)

## Groups

### Symmetric groups

#### Definition of odd and even permutations

$\sigma$ is an even permutation if the number of transpositions is even.

$\sigma$ is an odd permutation if the number of transpositions is odd.

#### Theorem for parity of transpositions

The parity of the number of transpositions is unique.

<details>
<summary>Proof</summary>

Prove using the determinant of a matrix, swapping the rows of the matrix multiply the determinant by $-1$.

Consider the identity matrix $I_n$. Then the determinant is $1$, let $(ij)A$, where $i\neq j$ denote the matrix obtained from $A$ by swapping the rows $j$ and $i$, then the determinant of $(1j)A$ is $-1$.

And,

$$
\det((a_1b_1)(a_2b_2)\cdots(a_nb_n)A)=(-1)^n\det(A)
$$

</details>

$S_3$ has 6 permutations $\{e,(12),(13),(23),(12)(23),(13)(23)\}$, 3 of them are even $\{e,(12)(23),(13)(23)\}$ and 3 of them are odd $\{(13),(12),(23)\}$.

#### Theorem for the number of odd and even permutations in symmetric groups

In general, $S_n$ has $n!$ permutations, half of them are even and half of them are odd.

<details>
<summary>Proof</summary>

Consider the set of odd permutations in $S_n$ and set of even permutations in $S_n$. Consider the function: $\alpha:S_n\to S_n$ where $\alpha(\sigma)=\sigma(12)$.

$\sigma$ is a bijection,

If $\sigma_1(12)=\sigma_2(12)$, then $\sigma_1=\sigma_2$.

If $\phi$ is an even permutation, $\alpha(\phi(12))=\phi(12)(12)=\phi$, therefore the number of elements in the set of odd and even permutations are the same.
</details>

#### Definition for sign of permutations

For $\sigma\in S_n$, the sign of $\sigma$ is defined by $\operatorname{sign}(\sigma)=1$ if sigma is even and $-1$ if sigma is odd.

Then $\beta: S_n\to \{1,-1\}$ is a group under multiplication, where $\beta(\sigma)=\operatorname{sign}(\sigma)$.

Then $\beta$ is a group homomorphism.

#### Definition of alternating group

$\ker(\beta)\leq S_n$, and $\ker(\beta)$ is the set of even permutations. Therefore the set of even permutations is a subgroup of $S_n$. We denote as $A_n$ (also called alternating group).

and $|A_n|=\frac{n!}{2}$.

### Direct product of groups

#### Definition of direct product of groups

Let $G_1,G_2$ be two groups. Then the direct product of $G_1$ and $G_2$ is defined as

$$
G_1\times G_2=\{(g_1,g_2):g_1\in G_1,g_2\in G_2\}
$$

The operations are defined by $(a_1,b_1)*(a_2,b_2)=(a_1*a_2,b_1*b_2)$.

This group is well defined since:

The identity is $(e_1,e_2)$, where $e_1\in G_1$ and $e_2\in G_2$. (easy to verify)

The inverse is $(a_1,b_1)^{-1}=(a_1^{-1},b_1^{-1})$.

Associativity automatically holds by associativity of $G_1$ and $G_2$.

<details>
<summary>Examples</summary>

Consider $\mathbb{Z}_\1\times \mathbb{Z}_2$.

$$
\mathbb{Z}_\1\times \mathbb{Z}_2=\{(0,0),(0,1),(1,0),(1,1)\}
$$

$(0,0)^2=(0,0)$, $(0,1)^2=(0,0)$, $(1,0)^2=(0,0)$, $(1,1)^2=(0,0)$

This is not a cyclic group, this is isomorphic to klein four group.

---

Consider $\mathbb{Z}_2\times \mathbb{Z}_3$.

$$
\mathbb{Z}_2\times \mathbb{Z}_3=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}
$$

This is cyclic ((2,3) are coprime)

Consider:

$$
\langle (1,1)\rangle=\{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2)\}
$$

</details>

#### Lemma for direct product of cyclic groups

$\mathbb{Z}_m\times \mathbb{Z}_n\simeq \mathbb{Z}_{mn}$ if and only if $m$ and $n$ have greatest common divisor $1$.

<details>
<summary>Proof</summary>

First assume $\operatorname{gcd}(m,n)=d>1$

Consider $(r,s)\in \mathbb{Z}_m\times \mathbb{Z}_n$.

We claim that order of $(r,s)$ is at most $\frac{mn}{d}<mn$.

Since $\frac{mn}{d}$ is integer, $\frac{mn}{d}=m_1dn_1$ where $m_1d$ is multiple of $m$ and $n_1d$ is multiple of $n$.

Therefore $r$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_m$ and $s$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_n$.

---

Other direction:

Assume $\operatorname{gcd}(m,n)=1$.

Claim order of $(1,1)=mn$, so $\mathbb{Z}_m\times \mathbb{Z}_n=\langle (1,1)\rangle$.

If $k$ is the order of $(1,1)$, then $k$ is a multiple of $m$ and a multiple of $n$.

</details>

Similarly, if $G_1,G_2,G_3,\ldots,G_k$ are groups, then

$$
G_1\times G_2\times G_3\times \cdots\times G_k=\{(g_1,g_2,\ldots,g_k):g_1\in G_1,g_2\in G_2,\ldots,g_k\in G_k\}
$$

is a group.

Easy to verify by associativity. $(G_1\times G_2)\times G_3=G_1\times G_2\times G_3$.

#### Some extra facts for direct product

1. $G_1\times G_2\simeq G_2\times G_1$, with $\phi(a_1,a_2)=(a_2,a_1)$.
2. If $H_1\leq G_1$ and $H_2\leq G_2$, then $H_1\times H_2\leq G_1\times G_2$.

> [!WARNING]
>
> Not every subgroup of $G_1\times G_2$ is of the form $H_1\times H_2$.
>
> Consider $\mathbb{Z}_2\times \mathbb{Z}_2$ with subgroup $\{(0,0),(1,1)\}$, This forms a subgroup but not of the form $H_1\times H_2$.