Section 15 Factor-Group Computations and Simple Groups

15.3 Examples:

• $G/\{0\}\simeq G$
• $G/G\simeq \{e\}$

NOTE

If G/N has order order of 2, then N must be normal.

15.4 Example

$|S_n|=2|A_n|$

• $A_n$:(Alternating group) is a normal subgroup of $S_n$

15.6 Falsity of the Converse of the Theorem of Lagrange

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=151&selection=47,13,64,1&color=note|p.146]]

It is false that if d divides the order of G, then there must exist a subgroup H of G having order d

$A_4$ (order 12) contains no subgroup of order 6.

15.7 Example

Compute $({\mathbb{Z}}_4\times {\mathbb{Z}}_6)/⟨(0,1)⟩$, where $⟨(0,1)⟩=\{(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)\}$

• The Cosets of H are: $(0,0)+H,(1,0)+H,(2,0)+H,(3,0)+H$
• It is clear that $({\mathbb{Z}}_4\times {\mathbb{Z}}_6)/⟨(0,1)⟩\simeq {\mathbb{Z}}_4$

Summary Let $G=H\times K$, then $G/\bar{H}\simeq K$, where $\bar{H}=\{(h,e)|h\in H\}$

15.9 Theorem

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A factor group of a cyclic group is cyclic.

15.11 Example

$({\mathbb{Z}}_4\times {\mathbb{Z}}_6)/⟨(2,3)⟩\simeq {\mathbb{Z}}_{12}$

G/H must have an order of 24/2=12 The possible abelian groups of order 12 are ${\mathbb{Z}}_4\times {\mathbb{Z}}_3$, and ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_3$

$({\mathbb{Z}}_4\times {\mathbb{Z}}_6)/⟨(2,3)⟩$ has an element of order 4, meaning ${\mathbb{Z}}_4\times {\mathbb{Z}}_3$ is the only possible answer, which is isomorphic to ${\mathbb{Z}}_{12}$ (Isomorphism of Cyclic Group)

15.12 Example

Compute $(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩$ Consider:

• each point as an Element of $\mathbb{Z}\times \mathbb{Z}$
• each $45^{\circ }$ lines as a coset of $\mathbb{Z}\times \mathbb{Z}$ or an Element of $(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩$
• Choose the y-axis as the representatives of the cosets
• The y-axis is isomorphic to $\mathbb{Z}$ $(\mathbb{Z}\times \mathbb{Z})/⟨(1,1)⟩\simeq \mathbb{Z}$

15.14 Definition Simple Groups

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A group is simple if it is nontrivial and has no proper nontrivial normal subgroups.

E.g. A group of prime order can have no nontrivial proper subgroups of any sort (The Theorem of Lagrange)

15.15 Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=154&selection=16,1,28,2&color=note|p.149]]

The alternating group $A_n$ is simple for n ≥ 5.

• Proof⏫

15.18 Theorem

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M is a maximal normal subgroup of G if and only if G/M is simple.

The Center and Commutator Sub Groups

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The Center and Commutator Subgroups

Skipped