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.

Proof

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)

$S_3$ 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.

Proof

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.

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 $G_2$ .

Examples

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)\}

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 $1$ .

Proof

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 $n$ .

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

$G_1\times G_2\simeq G_2\times G_1$ , with $\phi(a_1,a_2)=(a_2,a_1)$ .
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 $H_1\times H_2$ .