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. $gHg^{-1}=H$ for all $g\in G$
3. $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$.

$h\to ghg^{-1}$ 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 $G\trianglelefteq G$, that is, $i_{g}:G\to G$, where $i_{g}(a)=g\,a\,g^{-1}$ for all $a\in G$.

Automorphisms

Automorphisms of $G$ are of the form: $\phi :G\to G$, where $\phi$ 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 $g\,a\,g^{-1}=g\,b\,g^{-1}$, in particular $a=b$.
2. It is surjective: $i_{g}(g^{-1}\,a\,g)=a$. Thus if $i_{g}(b)=a$ if $b=g^{-1}\,a\,g$
3. $i_{g}$ is a homomorphism: $i_{g}(a\,b)=g\,a\,b\,g^{-1}=g\,a\,e\,b\,g^{-1}=(g\,a\,g^{-1})\,(g\,b\,g^{-1}$

Hence $i_{g}$ is an isomorphism.

Inner and Outer Automorphisms

If we have $f,g\in G$, then $i_{f}\circ i_{g}=f(g\,a\,g^{-1})f^{-1}=(f\,g)\,a\,(f\,g)^{-1}=id_{f\,g}$.

The group $\mathrm {Inn} \,G=\left\{i_{g}~\mid ~g\in G\right\}$ is a subgroup of $\mathrm {Aut} (G)$ 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. $gHg^{-1}=H$ for all $g\in G$ if and only if $i_{g}[H]=G$
3. $ghg^{-1}\in H$ 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 $H\trianglelefteq G$, $L\trianglelefteq L_{1}$, and $K\trianglelefteq K_{1}$, 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 $a\neq e$.

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

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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 $A_{n}\triangleleft S_{n}$ because we can express $S_{n}$ as the disjoint union of $A_{n}$ and $\tau \,A_{n}$ ($S_{n}=A_{n}\sqcup \tau \,A_{n}$) for any $\tau \not \in A_{n}$.

Equivalently, $\left|S_{n}/A_{n}\right|=2$, so $S_{n}/A_{n}$ is isomorphic to $\mathbb {Z} _{2}$.

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Let $G=\mathbb {Z} _{4}\times \mathbb {Z} _{6}$ is abelian, and let $H=\left\langle (0,1)\right\rangle$ be the cyclic subgroup generated by $(0,1)$. Note that $H\triangleleft G$.

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

$|H|=6$ because $H=\left\{(0,0),(0,1),\ldots ,(0,5)\right\}$

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

This example calls to mind that any group of the form $H=\left\{0\right\}\times \mathbb {Z} _{6}$, so $G/H=\mathbb {Z} _{4}/\left\{0\right\}\times \mathbb {Z} _{6}/\mathbb {Z} _{6}\simeq \mathbb {Z} _{4}\times \left\{0\right\}\simeq \mathbb {Z} _{4}$.

What about $H=\left\langle (2,3)\right\rangle$? Then $H=\left\{(2,3),(0,0)\right\}$, so $H\simeq \mathbb {Z} _{2}$. Then $\left|G/H\right|={\frac {24}{2}}=12$.

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. $\mathbb {Z} _{2}\times \mathbb {Z} _{2}\times \mathbb {Z} _{3}$
2. $\mathbb {Z} _{4}\times \mathbb {Z} _{3}$

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 $\mathbb {Z} _{4}\times \mathbb {Z} _{3}$.

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 $G=\left\langle a\right\rangle$ and $H=\left\langle a^{s}\right\rangle$, then $G/H$ is also cyclic.

Proof. $G/H=\left\langle aH\right\rangle =\left\{H,aH,a^{2}H\right\}$. 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 $G=\mathbb {Z} \times \mathbb {Z}$ and $H=\left\langle (1,1)\right\rangle$. 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, $G/H\simeq \mathbb {Z}$.

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: $e_{G}\in H$, $a\,b\in H$ for $a,b\in H$, and $a\in H$ implies $a^{-1}\in H$
• Cyclic Groups and Subgroups: groups that can be generated by $a\in G$: $G=\left\langle a\right\rangle =\left\{a^{n}~\mid ~n\in \mathbb {Z} \right\}$, where $a^{n}=e$
  • Orders of elements: $\mathrm {order} (a)=n$ (or $\infty$ if $n$ does not exist
  • Note that the order of $a$ (or the order of $\left\langle a\right\rangle$) is denoted $\mathrm {order} (a)=\left|\left\langle a\right\rangle \right|$, not $|a|$!
  • In all cases, either $\left\langle a\right\rangle <G$ is a proper subgroup or $\left\langle a\right\rangle =G$ generates the group.
  • Cyclic subgroups are isomorphic to $\mathbb {Z}$ if infinite, and isomorphic to $\mathbb {Z} _{n}$ 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 $n\mathbb {Z}$
  • By Theorem 6.14, $H<\mathbb {Z} _{n}$
  • 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