MATH 415 Lecture 12

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End Exam 1 content

Exam on Tuesday.

Group Factorization

Normal Subgroups

Normal Subgroups

Any of the following conditions are equivalent:

1. $aH=Ha$ for all $a\in G$.
2. $gH{g}^{-1}=H$ for all $g\in G$
3. $gh{g}^{-1}\in H$ for all $g\in G$ and $h\in H$.

$h\to gh{g}^{-1}$ is called the conjugation of $h$.

Let's call this function $i_g:H\to H$. It happens that this function also works over $G$ (since $G⊴G$, that is, $i_g:G\to G$, where $i_g(a)=ga{g}^{-1}$ for all $a\in G$.

Automorphisms

Automorphisms of $G$ are of the form: $\varphi :G\to G$, where $\varphi$ is an isomorphism. The set of Automorphisms is a group over composition

Theorem. From our example above, $i_g$ is a member of the automorphisms of $G$:

1. It is injective: $i_g(a)=i_g(b)$ implies $ga{g}^{-1}=gb{g}^{-1}$, in particular $a=b$.
2. It is surjective: $i_g(g^{-1}ag)=a$. Thus if $i_g(b)=a$ if $b=g^{-1}ag$
3. $i_g$ is a homomorphism: $i_g(ab)=gab{g}^{-1}=gaeb{g}^{-1}=(ga{g}^{-1})(gb{g}^{-1}$

Hence $i_g$ is an isomorphism.

Inner and Outer Automorphisms

If we have $f,g\in G$, then $i_f\circ i_g=f(ga{g}^{-1})f^{-1}=(fg)a(fg{)}^{-1}=i{d}_{fg}$.

The group $\mathrm{I}\mathrm{n}\mathrm{n}G=\{i_g\mid g\in G\}$ is a subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(G)$ with identity element $i_e=id$. We call this group the inner automorphisms of $G$

Inn(G) is a normal subgroup of Aut(G), so Aut(G)/Inn(G) exists, and we wil call this group the outer automorphisms of G

Going batk to our above criteria, we add the following:

1. $aH=Ha$ for all $a\in G$.
2. $gH{g}^{-1}=H$ for all $g\in G$ if and only if $i_g[H]=G$
3. $gh{g}^{-1}\in H$ for all $g\in G$ and $h\in H$.
4. $H$ is invariant with respect to inner automorphisms

For a group $G$, we can construct $G$ as the disjoint union of all left cosets $aH$, where $a\in T$ for transversal $N$.

It was mentioned last time that

• $G/\left\{e\right\}\simeq G$
• $G/G\simeq \left\{e\right\}$
• If $G\times L\times K$ and $H⊴G$, $L⊴L_1$, and $K⊴K_1$, then $G/H\simeq L/L_1\times K/K_1$

If $H<G$, then the index $(G:H)=\frac{\left|G\right|}{\left|H\right|}$ is equivalent to the number of cosets.

• If $G$ is finite, then the index must be finite.
• If $G$ is infinite, then the index may or may not be finite.

In the case where $(G:H)=2$, then Failed to parse (unknown function "\lefttriangle"): {\displaystyle H \lefttriangle G} is normal. In particular, we have $eH$ and $aH$, where $a\ne e$.

This does not hold necessarily for $(G:H)=3$, etc.

---

Let $G=S_n$ be the symmetric group, and let $H=A_n$ be the alternating group. By definition, $\sigma \in A_n$ if and only if $\sigma$ can be expressed as a product of an even number of transpositions.

Therefore $S_n\setminus A_n$ consists of all odd permutations, and $(S_n:A_n)=2$. Thus $A_n◃S_n$ because we can express $S_n$ as the disjoint union of $A_n$ and $\tau A_n$ ($S_n=A_n\bigsqcup \tau A_n$) for any $\tau \in ̸A_n$.

Equivalently, $|S_n/A_n|=2$, so $S_n/A_n$ is isomorphic to ${\mathbb{\mathbb{Z}}}_2$.

---

Let $G={\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_6$ is abelian, and let $H=⟨(0,1)⟩$ be the cyclic subgroup generated by $(0,1)$. Note that $H◃G$.

$|G|=26$ because ${\mathbb{\mathbb{Z}}}_4$ and ${\mathbb{\mathbb{Z}}}_6$ have 4 and 6 elements respectively.

$|H|=6$ because $H=\{(0,0),(0,1),\dots ,(0,5)\}$

Therefore $|G/H|=\frac{|G|}{|H|}=\frac{24}{6}=4$.

This example calls to mind that any group of the form $H=\left\{0\right\}\times {\mathbb{\mathbb{Z}}}_6$, so $G/H={\mathbb{\mathbb{Z}}}_4/\left\{0\right\}\times {\mathbb{\mathbb{Z}}}_6/{\mathbb{\mathbb{Z}}}_6\simeq {\mathbb{\mathbb{Z}}}_4\times \left\{0\right\}\simeq {\mathbb{\mathbb{Z}}}_4$.

What about $H=⟨(2,3)⟩$? Then $H=\{(2,3),(0,0)\}$, so $H\simeq {\mathbb{\mathbb{Z}}}_2$. Then $|G/H|=\frac{24}{2}=12$.

We know that this group is abelian by the #Theorem, so it is isomorphic to one of the following groups (only choices for abelian groups of order 12):

1. ${\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_2\times {\mathbb{\mathbb{Z}}}_3$
2. ${\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_3$

The main difference between the first group and the second is that the first has no element of order 4. In fact, its only maximal orders are 2, 3, and 2×3=6.

Therefore, our group $G/H$ must be isomorphic to ${\mathbb{\mathbb{Z}}}_4\times {\mathbb{\mathbb{Z}}}_3$.

Theorem

Theorem. A factor of an abelian group is also abelian: Given an abelian group $G$ and $H<G$, then $G/H$ is abelian.

Proof. We can find that

$(aH)(bH)=(bH)(aH)$

in

$G/H$

because

$(ab)H=(ba)H$

, which is true by the fact that

$G$

is abelian.

quod erat demonstrandum

Theorem. A factor group of a cyclic group is also cyclic. Given $G=⟨a⟩$ and $H=⟨a^s⟩$, then $G/H$ is also cyclic.

Proof.

$G/H=⟨aH⟩=\{H,aH,a^{2}H\}$

. Observe that

$\left\{a^n\right\}=G$

, so

$G/H$

can be expressed in cyclic form generated by the coset containing

$a$

.

quod erat demonstrandum

Example 5.12

Let $G=\mathbb{\mathbb{Z}}\times \mathbb{\mathbb{Z}}$ and $H=⟨(1,1)⟩$. What is $G/H$?

Semantically, $G$ represents the set of points with integer coordinates, and $H$ represents the points along the diagonal $y=x$.

The cosets $aH$ of $G$ will represent the diagonal lines that are parallel to $H$ with integer points.

The transversal can be formed from any line perpendicular to the cosets. Therefore the index $(G:H)$ is infinite, and its quotient is infinite. In particular, $G/H\simeq \mathbb{\mathbb{Z}}$.

Exam Review

Bring your own paper.

Calculator is not needed, but if wanted, only 4-function

Start with name.

• Complex Numbers, roots of unity, etc
• Binary Structures: set with binary operation
  • Isomorphism of binary structure
  • Identity element of binary structure
• Groups and Subgroups (axioms and definitions)
  • Def of Subgroup: $e_G\in H$, $ab\in H$ for $a,b\in H$, and $a\in H$ implies $a^{-1}\in H$
• Cyclic Groups and Subgroups: groups that can be generated by $a\in G$: $G=⟨a⟩=\{a^n\mid n\in \mathbb{\mathbb{Z}}\}$, where $a^n=e$
  • Orders of elements: $\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}(a)=n$ (or $\mathrm{\infty }$ if $n$ does not exist
  • Note that the order of $a$ (or the order of $⟨a⟩$) is denoted $\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}(a)=\left|⟨a⟩\right|$, not $|a|$!
  • In all cases, either $⟨a⟩<G$ is a proper subgroup or $⟨a⟩=G$ generates the group.
  • Cyclic subgroups are isomorphic to $\mathbb{\mathbb{Z}}$ if infinite, and isomorphic to ${\mathbb{\mathbb{Z}}}_n$ if finite and $\left|G\right|=n$.
  • A subgroup of cyclic group is also cyclic.
  • Subgroups of cyclic groups can always be expressed as $\left\{0\right\}$ or $n\mathbb{\mathbb{Z}}$
  • By Theorem 6.14, $H<{\mathbb{\mathbb{Z}}}_n$
  • If $G$ is a finite cyclic group generated by $a$, and $H<G$ is a cyclic subgroup generated by $a^s$, then for $d=\gcd (s,n)$, $\left|H\right|=\frac{n}{d}$.
• Permutations, symmetric group $S_n$ from all permutations of $n$ elements, and alternating group $A_m$ from all even permutations
  • cycles, orbits, etc.
• Direct Products
  • Order of element of direct product is LCM of each element in respective group.
• Fundamental Theorem of Finitely Generated Abelian Groups
• Normal Subgroups (definitions)
  • Homomorphism, Kernel
• Factor Groups
  • Definition of product of cosets