NoteNextra-origin/content/Math4302/Math4302_L11.md

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.

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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.

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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

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.