eead5b6458
3.4 KiB
Math4302 Modern Algebra (Lecture 6)
Subgroups
Dihedral group
The dihedral group D_n is the group of all rotations and reflections about the center of the regular polygon of n sides.
|S_n|=n!, |D_n|=2n
Cyclic group
G=\langle a\rangle=\{1,a,a^2,\cdots\} for some a\in G
Example of cyclic group
(\mathbb{Z}_n,+) is cyclic and \mathbb{Z}_n=\langle 1\rangle=\{0,1,2,\cdots,n-1\}
(\mathbb{Z},+) is cyclic and \mathbb{Z}=\langle 1\rangle=\langle -1 \rangle
S_3 is not cyclic
\langle e\rangle=\{e\}
\langle (1,2)\rangle=\{e,(1,2)\}
\langle (1,3)\rangle=\{e,(1,3)\}
\langle (2,3)\rangle=\{e,(2,3)\}
\langle (1,2,3)\rangle=\{e,(1,2,3),(1,3,2)\}
\langle (1,3,2)\rangle=\{e,(1,3,2),(1,2,3)\}
Every cyclic group is abelian
Every cyclic group is abelian
Proof
Let G=\langle a\rangle be a cyclic group, then \forall g_1,g_2\in G we have g_1g_2=g_2g_1 since g_1g_2=a^k_1a^k_2=a^{k_1+k_2} and g_2g_1=a^k_2a^k_1=a^{k_1+k_2}
Definition for order of element
Let G be a group, then the order of g\in G is defined to be the size of the smallest subgroup containing g.
If |\langle g\rangle| is infinite, then we say that g has infinite order.
Example of order of element
5 in (\mathbb{Z},+) has infinite order.
5 in (\mathbb{Z}_{10},+) has order 2.
\langle 5\rangle=\{0,5\}.
5 in (\mathbb{Z}_{6},+) has order 6.
\langle 5\rangle=\{0,5,4,3,2,1\}.
Lemma for order of element
Let G be a group, then a\in G has order n if n is the smallest positive integer such that a^n=e.
Proof
There are 2 cases:
Case 1:
There is no positive n such that a^n=e.
Then a^i\neq a^j if i\neq j, i,j\in \mathbb{N}.
Reason: if a^i=a^j, then a^{i-j}=e.
Then the order of group is infinite.
Case 2:
There is a positive n such that a^n=e.
Let n be the smallest such positive integer. Then we claim \langle a^n\rangle=\{e,a^1,a^2,\cdots,a^{n-1}\}.
We claim they are all distinct.
Suppose not, then we can have a^i=a^j for i\neq j, 0\leq i,j\leq n-1.
Then a^{i-j}=e but i-j\leq n-1. Therefore n is not the smallest positive integer such that a^n=e.
Theorem for cyclic group up to isomorphism
Suppose G is a cyclic group,
- If
|G|=n, then|G|\simeq \mathbb{Z}_n^+ - If
|G|=\infty, then|G|\simeq \mathbb{Z}.
Proof
Case 1:
If |G|=\infty, then we can map G to (\mathbb{Z},+), where G=\langle a\rangle. \phi(n)=a^n. This gives a bijection between G and (\mathbb{Z},+).
where \phi(n+m)=a^{n+m}=a^n a^m=\phi(n)\phi(m).
Case 2:
If |G|=n, then we can map G to (\mathbb{Z}_n,+), where G=\langle a\rangle. \phi(n)=a^n. This gives a bijection between G and (\mathbb{Z}_n,+).
where \phi(n+m)=\phi(r)=a^{n+m}=a^n a^m=\phi(n)\phi(m).
Example
Let H=\langle (12)(345)\rangle\subseteq S_5. Then H\simeq \mathbb{Z}_6^+.
Let \tau=(12)(345)
All the elements of H are:
\tau^0=(12)(345)\tau^1=(453)\tau^2=(12)(534)\tau^3=(345)\tau^4=(12)(453)\tau^5=(534)
GCD and order
If G=\langle a\rangle, then H=\langle a^k\rangle, |H|=\frac{n}{d} where d=\operatorname{gcd}(n,k).