mth3003 portfolio cheat sheet

MTH3003 Group Theory - Portfolio Cheat Sheet

Created by William Fayers. Good luck and have fun!! :))

0. Reference Tables & Foundations

Definitions

Concept	Definition
Group $(G,\ast )$	Closure + Identity $e_G$ + Inverses + Associativity
Subgroup $H\le G$	$H\subseteq G$, itself a group under same operation
Normal $N⊴G$	$g^{-1}kg\in N\forall k\in N,g\in G$. Equiv: $gN=Ng\forall g$. Equiv: $g^{-1}Ng=N\forall g$
Coset $gH$	$\{gh:h\in H\}$. Index $[G:H]$ = number of distinct cosets
Quotient $G/N$	$\{gN:g\in G\}$ with $(aN)(bN)=(ab)N$. Requires $N⊴G$
Homomorphism $\theta :G\to H$	$\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)\forall g_1,g_2$
Kernel / Image	$\text{Ker}(\theta )=\{g:\theta (g)=e_H\}$, $\text{Im}(\theta )=\{\theta (g):g\in G\}$
Isomorphism	Bijective homomorphism. $G\cong H$
$o(g)$, $\text{mod}(G)$	$o(g)$ = smallest $n$ with $g^n=e$. $\text{mod}(G)$ = number of elements
Cyclic $⟨g⟩$	$\{g^n:n\in \mathbb{Z}\}$. $G$ cyclic if $G=⟨g⟩$. $\text{mod}(⟨g⟩)=o(g)$
Abelian	$ab=ba\forall a,b$. Simple: only normal subgroups are $\{e\}$ and $G$
$\text{FS}(X)$	$\{g\in \text{Sym}(X):\text{supp}(g)=\{x:g(x)\ne x\}\text{ is finite}\}$

Note: $\text{mod}(G)=|G|$, just used different notation otherwise it’d break the table.

Quick Subgroup Test (Theorem 2.2.5)

$H\le G⟺$ (i) $e_G\in H$, (ii) $h_1,h_2\in H\Rightarrow h_1{h}_2\in H$, (iii) $h\in H\Rightarrow h^{-1}\in H$.

Basic Group Theorems (Theorem 2.1.11)

• $(g^{-1}{)}^{-1}=g$. Cancellation: $g{h}_1=g{h}_2\Rightarrow h_1=h_2$.
• Socks-and-shoes: $(h_1{h}_2{)}^{-1}=h_2^{-1}{h}_1^{-1}$. Proof: $(h_1{h}_2)(h_2^{-1}{h}_1^{-1})=e$.
• Identity and inverses are unique (prove by cancellation).

Subgroup/Equality Facts

• $|H|=1⟺H=\{e\}$. Finite: $|H|=|G|⟺H=G$ (since $H\subseteq G$). Infinite: FAILS ($\mathbb{Z}<\mathbb{Q}$ but $|\mathbb{Z}|=|\mathbb{Q}|$).

Normal Subgroup Equivalences

$(i)⟺(iii)$: $gN\subseteq Ng$ since $gk=(gk{g}^{-1})g\in Ng$; and $Ng\subseteq gN$ since $kg=g(g^{-1}kg)\in gN$. So $gN=Ng$.

Important Groups

Group	Notation	Order	Notes
Symmetric	$S_n$	$n!$	All permutations of $\{1,\dots ,n\}$
Alternating	$A_n$	$n!/2$	Even perms; $A_n⊴S_n$; simple for $n\ge 5$
Cyclic	$C_n=⟨(12\dots n)⟩$	$n$	Single generator
Dihedral	$D_{2n}$	$2n$	Symmetries of regular $n$-gon
Klein four	$K_4$	4	$\{e,(12),(34),(12)(34)\}$; smallest non-cyclic
${\mathbb{Z}}_n$	$⟨[1{]}_n⟩$	$n$	Integers mod $n$ under $\oplus$
${\text{GL}}_2(\mathbb{R})$	-	$\infty$	Invertible $2\times 2$ matrices ($\det \ne 0$)
${\text{SL}}_2(\mathbb{R})$	-	$\infty$	$\det =1$
$P$	-	$\infty$	$\det >0$
$\text{FS}(\mathbb{Z})$	-	$\infty$	Finitary perms of $\mathbb{Z}$

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1. Permutations

1a. Converting Representations

Two-row → cycles: Start at smallest unused element, follow $\sigma$ until you return. Omit fixed points. Cycles → two-row: In $(a_1{a}_2\dots a_r)$: $a_i\to a_{i+1}$, $a_r\to a_1$. Unlisted elements are fixed.

1b. Composition

$\sigma \rho$ means “apply $\rho$ first, then $\sigma$“. Chase each element: $x\to \sigma (\rho (x))\to \cdots$ until cycle closes. Disjoint cycles commute.

1c. Inversion

Reverse each cycle: $(a_1{a}_2\dots a_m{)}^{-1}=(a_m\dots a_2{a}_1)$. For disjoint product: invert each factor (order doesn’t matter).

1d. Order

Single $r$-cycle has order $r$. Disjoint cycles of lengths $r_1,\dots ,r_m$: $o(\sigma )=\text{lcm}(r_1,\dots ,r_m)$.

• Proof: Disjoint cycles commute, so ${\sigma }^k=e$ if and only if $r_i\mid k\forall i$ if and only if $\text{lcm}\mid k$.
• Find order $d$ in $S_n$: choose cycle lengths summing to $\le n$ with lowest common multiplier $=d$.
• No order $p$ (prime $>n$): would need a $p$-cycle, requiring $p\le n$.

1e. Solving Equations

$\alpha \sigma =\beta \Rightarrow \sigma ={\alpha }^{-1}\beta$. $\sigma \alpha =\beta \Rightarrow \sigma =\beta {\alpha }^{-1}$. $\alpha \sigma \gamma =\beta \Rightarrow \sigma ={\alpha }^{-1}\beta {\gamma }^{-1}$. Always verify.

1f. Foundational Results

Disjoint cycles commute (Prop 1.2.7): If $\{a_i\}\cap \{b_j\}=\varnothing$, then for $x=a_i$: both $\sigma \rho (x)$ and $\rho \sigma (x)$ give $a_{i+1}$ (since $\rho$ fixes $a_{i+1}$). Similarly for $b_j$‘s and fixed points.

Disjoint cycle decomposition (Prop 1.3.4): Start at $x_1$, follow $\sigma$ until first repeat - must be $x_1$ (by injectivity), giving a cycle. Remove and repeat on remaining elements.

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2. Subgroups, Normality & Cyclic Groups

2a. Proving $H\le G$

Apply Quick Subgroup Test. Key examples:

• $H\cap K\le G$: $e\in H\cap K$; $a,b\in H\cap K\Rightarrow ab\in H$ and $ab\in K$; $a^{-1}\in H$ and $a^{-1}\in K$.
• $\text{Im}(\theta )\le H$: $e_H=\theta (e_G)\in \text{Im}$; $\theta (g_1)\theta (g_2)=\theta (g_1{g}_2)\in \text{Im}$; $\theta (g{)}^{-1}=\theta (g^{-1})\in \text{Im}$.
• $\text{FS}(\mathbb{Z})\le \text{Sym}(\mathbb{Z})$: $\text{supp}(e)=\varnothing$; $\text{supp}(fg)\subseteq \text{supp}(f)\cup \text{supp}(g)$; $\text{supp}(g^{-1})=\text{supp}(g)$.
• $A_n\le S_n$: $\sigma (e)=1$; $\sigma (gh)=\sigma (g)\sigma (h)=1$; $\sigma (g^{-1})=\sigma (g)=1$.

2b. Proving $N⊴G$

Show $g^{-1}kg\in N\forall k\in N,g\in G$. Shortcuts:

• $G$ Abelian $\Rightarrow$ every subgroup normal ($g^{-1}kg=k$).
• $N=\text{Ker}(\theta )\Rightarrow N⊴G$ automatically.
• $[G:N]=2\Rightarrow N⊴G$ (only two cosets, so $gN=Ng$).
• Determinant trick: $\det (M^{-1}AM)=\det (A)$, so $\text{det}$-defined subgroups (${\text{SL}}_2$, $P$) are normal.
• Support trick: $\text{supp}(g^{-1}hg)\subseteq g^{-1}(\text{supp}(h))$, same finite size. So $\text{FS}(X)⊴\text{Sym}(X)$.

Examples: $A_n⊴S_n$: $\sigma (g^{-1}hg)=\sigma (g{)}^{-1}\cdot 1\cdot \sigma (g)=1$. $C_3⊴S_3$: check by cases. $H\cap N⊴H$ (when $N⊴G$): $h^{-1}kh\in H$ (closure) and $h^{-1}kh\in N$ (normality).

2c. Cyclic Groups

$G=⟨g⟩$ if every element is a power of $g$. For $|G|=p$ prime: any $g\ne e$ has $|⟨g⟩|$ dividing $p$, so $|⟨g⟩|=p=|G|$. For ${\mathbb{Z}}_n$: $[m{]}_n=m\cdot [1{]}_n$.

2d. Listing $⟨g⟩$

Compute $e,g,g^2,\dots$ until return to $e$. Example: $g=(1357)(24)\in S_{10}$, $o(g)=4$. $⟨g⟩=\{e,g,(15)(37),(1753)(24)\}$. Subgroup: $⟨g^2⟩=\{e,(15)(37)\}$.

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3. Lagrange’s Theorem

Statement (Theorem 4.2.5)

$H\le G$ finite $\Rightarrow |G|=[G:H]\cdot |H|$. So $|H|\mid |G|$ and $o(g)\mid |G|$.

Proof: $G$ is a disjoint union of $[G:H]$ cosets (Cor 4.1.4), each of size $|H|$ (bijection $h\mapsto gh$).

$o(g)\mid |G|$: $o(g)=|⟨g⟩|$, and $⟨g⟩\le G$, so Lagrange applies.

Applications

• No subgroup/element of order $k$ if $k\nmid |G|$. Converse fails: $A_4$ (order 12) has no subgroup of order 6.
• Cauchy: $p$ prime, $p\mid |G|\Rightarrow \exists$ element of order $p$.
• $C_p$ is simple: subgroup orders divide $p$, so only $\{e\}$ and $C_p$.

Cosets

• $aH=bH⟺a^{-1}b\in H$. Proof ($\Rightarrow$): $a\in bH\Rightarrow a=bh\Rightarrow a^{-1}b=h^{-1}\in H$. ($\Leftarrow$): $a^{-1}b=h_0\Rightarrow ah=b(h_0^{-1}h)\in bH$ so $aH\subseteq bH$; symmetrically $bH\subseteq aH$.
• Cosets partition $G$: if $aH\cap bH\ni g=a{h}_1=b{h}_2$, then $a^{-1}b=h_1{h}_2^{-1}\in H$, so $aH=bH$.

Listing cosets: Start with $eH$; pick $g\notin H$, form $gH$; repeat until $G$ exhausted. Example: $H=\{e,(12)\}$ in $S_3$: cosets $\{e,(12)\}$, $\{(123),(13)\}$, $\{(132),(23)\}$. $[S_3:H]=3$.

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4. Dihedral Groups $D_{2n}$

Structure & Relations

$D_{2n}=\{e,\rho ,\dots ,{\rho }^{n-1},\sigma ,\sigma \rho ,\dots ,\sigma {\rho }^{n-1}\}$, $|D_{2n}|=2n$.

• $\rho =(12\dots n)$ (rotation by $2\pi /n$), $\sigma =(2n)(3n-1)\cdots$ (reflection through vertex 1).
• ${\rho }^n=e$, ${\sigma }^2=e$, ${\rho }^k\sigma =\sigma {\rho }^{-k}=\sigma {\rho }^{n-k}$.
• Proof: $(\rho \sigma {)}^2=e$ (reflection has order 2), so $\rho \sigma ={\sigma }^{-1}{\rho }^{-1}=\sigma {\rho }^{-1}$.

Computing in $D_{2n}$

Push $\sigma$‘s left using $\rho \sigma =\sigma {\rho }^{-1}$, reduce with ${\sigma }^2=e$, ${\rho }^n=e$. Example: ${\rho }^3\sigma =\sigma {\rho }^{-3}=\sigma {\rho }^{n-3}$.

Cycle Notation for $D_{2n}$

Rotations: ${\rho }^k$ sends $i\to i+k(\mathrm{mod}n)$. Reflections: draw $n$-gon, track vertices geometrically.

$D_{10}$: $\rho =(12345)$, $\sigma =(25)(34)$. $\sigma \rho =(15)(24)$, $\sigma {\rho }^2=(14)(23)$, $\sigma {\rho }^3=(13)(45)$, $\sigma {\rho }^4=(12)(35)$.

Symmetry Groups & Construction

Label vertices, find all rotations and reflections as permutations. Regular $n$-gon $\to D_{2n}$. Cross/plus with 8 corners: 4 rotations + 4 reflections $=D_8$. To build $D_{2n}$: regular $n$-gon with matching decorations.

$⟨\rho ,\sigma ⟩=D_{2n}$

Any word reduces to ${\sigma }^k{\rho }^j$ via $\rho \sigma =\sigma {\rho }^{-1}$. Since ${\sigma }^2=e$, ${\rho }^n=e$: $k\in \{0,1\}$, $j\in \{0,\dots ,n-1\}$, giving $2n$ elements.

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5. Homomorphisms & Isomorphisms

5a. Proving/Disproving Homomorphism

Check $\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)\forall g_1,g_2$. To disprove: one counterexample suffices.

• $\varphi (g)=g^2$: homomorphism if and only if $G$ Abelian (since $(g_1{g}_2{)}^2=g_1^2{g}_2^2$ if and only if $g_1{g}_2=g_2{g}_1$).
• Canonical map $\varphi (h)=hN$: $\varphi (h_1{h}_2)=h_1{h}_{2}N=(h_{1}N)(h_{2}N)=\varphi (h_1)\varphi (h_2)$.
• Canonical map $\pi :G\to G/N$, $\pi (g)=gN$: surjective homomorphism with $\text{Ker}(\pi )=N$.

5b. Properties (Prop 5.2.4)

$\theta :G\to H$ homomorphism $\Rightarrow$: $\theta (e_G)=e_H$ (cancel $\theta (g)$ from $\theta (g)=\theta (e_{G}g)$). $\theta (g^{-1})=\theta (g{)}^{-1}$. $\theta (g^m)=\theta (g{)}^m$ (induction). $\text{Im}(\theta )\le H$. $\text{Ker}(\theta )⊴G$.

5c. Proving Isomorphism

Construct $\theta$ and verify: (1) homomorphism, (2) injective ($\text{Ker}=\{e\}$), (3) surjective. If $|G|=|H|$ finite: injective $⟺$ surjective.

• Power maps: $g\mapsto g^k$ on $C_n$ is iso if and only if $\gcd (k,n)=1$.
• Strategy: define on generators, extend. E.g. $D_6\cong S_3$: $\theta (\rho )=(123)$, $\theta (\sigma )=(12)$.

5d. $\text{Ker}(\theta )⊴G$

$\theta (g^{-1}ag)=\theta (g{)}^{-1}\theta (a)\theta (g)=\theta (g{)}^{-1}{e}_H\theta (g)=e_H$, so $g^{-1}ag\in \text{Ker}(\theta )$.

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6. Isomorphism Theorems

First (Theorem 6.0.1): $G/\text{Ker}(\varphi )\cong \text{Im}(\varphi )$

Via $\theta (gK)=\varphi (g)$. Use: find $\text{Ker}$ and $\text{Im}$ of a homomorphism, conclude quotient $\cong$ image.

Proof: (1) Well-defined: $g_{1}K=g_{2}K\Rightarrow g_1^{-1}{g}_2\in K\Rightarrow \varphi (g_1)=\varphi (g_2)$. (2) Homomorphism: $\theta (g_{1}K{g}_{2}K)=\varphi (g_1{g}_2)=\varphi (g_1)\varphi (g_2)$. (3) Onto: $\varphi (g)=\theta (gK)$. (4) 1-1: $\theta (g_{1}K)=\theta (g_{2}K)\Rightarrow \varphi (g_1)=\varphi (g_2)\Rightarrow g_1^{-1}{g}_2\in K$.

Applications: $\mathbb{Z}/n\mathbb{Z}\cong {\mathbb{Z}}_n$; $S_n/A_n\cong C_2$ (signature); ${\text{GL}}_2/{\text{SL}}_2\cong {\mathbb{R}}^{\ast }$ (det); $P/{\text{SL}}_2\cong {\mathbb{R}}_{>0}$; ${\text{GL}}_2/P\cong C_2$ (sign of the determinant, $\text{sgn det}$).

Second (Theorem 6.0.4): $H/(H\cap N)\cong (HN)/N$

Map $\varphi (h)=hN$: homomorphism with $\text{Ker}=H\cap N$. Apply FIT.

Third (Theorem 6.0.5): $(G/N)/(M/N)\cong G/M$

(“Fool’s Cancellation.“) Proof: $\varphi (gN)=gM$. Well-defined: $g_1^{-1}{g}_2\in N\subseteq M$. Homomorphism: yes. Onto: yes. $\text{Ker}=M/N$. Apply FIT.

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7. Alternating Group & Signature

Signature Function

$\sigma (g)={\prod }_i(-1{)}^{r_i-1}$ where $r_i$ are disjoint cycle lengths. Even: $\sigma =1$. Odd: $\sigma =-1$. Quick rule: count even-length cycles; odd count $\to$ odd perm.

• $\sigma$ is a homomorphism: $\sigma (gh)=\sigma (g)\sigma (h)$. Also $\sigma (g^{-1})=\sigma (g)$ and $\sigma (g^{-1}hg)=\sigma (h)$.

Proof $\sigma$ is homomorphism (Prop 7.2.2): $\Delta ={\prod }_{i<j}(x_i-x_j)$. $gh$ sends $\Delta \to g(\sigma (h)\Delta )=\sigma (h)\sigma (g)\Delta$, so $\sigma (gh)=\sigma (g)\sigma (h)$.

$A_n=\text{Ker}(\sigma )$

$A_n\le S_n$ (Quick Subgroup Theorem: $\sigma (e)=1$, $\sigma (gh)=1\cdot 1=1$, $\sigma (g^{-1})=1$). $A_n⊴S_n$ (kernel of homomorphism). $|A_n|=n!/2$ (FIT: $S_n/A_n\cong C_2$ since $\sigma$ surjective). Simple for $n\ge 5$.

Cycle Shape Tables

$S_3$: $\varnothing$(+1,1), $(2)$(-1,3), $(3)$(+1,2). $A_3=\{e,(123),(132)\}$.

$S_4$: $\varnothing$(+1,1), $(2)$(-1,6), $(2,2)$(+1,3), $(3)$(+1,8), $(4)$(-1,6). $|A_4|=12$. Note: $K_4⊴A_4$, so $A_4$ not simple.

$S_5$: $\varnothing$(+1,1), $(2)$(-1,10), $(2,2)$(+1,15), $(3)$(+1,20), $(3,2)$(-1,20), $(4)$(-1,30), $(5)$(+1,24). $|A_5|=60$.

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8. Proof Toolbox

Order Switching Lemma (Lemma 4.2.3)

$N⊴G$, $g\in G$, $k\in N\Rightarrow \exists k^{\prime },k^{\prime \prime }\in N$: $gk=k^{\prime }g$ and $kg=g{k}^{\prime \prime }$. (Since $gN=Ng$.)

Key Lemma (Lemma 4.2.4): $H\le G$, $N⊴G$

$HN=NH$ (Order Switching). $H\cap N⊴H$. $HN\le G$ (Quick Subgroup Theorem + Order Switching for closure/inverses).

More Tools

• Conjugation power: $(g^{-1}hg{)}^n=g^{-1}{h}^{n}g$ (expand, cancel $g{g}^{-1}$ pairs).
• Division algorithm: $o(g)=n\Rightarrow ⟨g⟩=\{e,g,\dots ,g^{n-1}\}$. Any $g^m=g^r$ where $m=qn+r$.
• Abelian: every subgroup normal; $g\mapsto g^k$ always a homomorphism; $(gh{)}^n=g^n{h}^n$.
• Non-Abelian: $(gh{)}^2\ne g^2{h}^2$ in general. $\text{FS}(\mathbb{Z})$ is non-Abelian (contains $S_n$ for all $n$).

$G/N$ Is a Well-Defined Group

• Well-defined: $a_{1}N=a_{2}N$, $b_{1}N=b_{2}N$ with $a_2=a_1{k}_1$, $b_2=b_1{k}_2$. Then $a_2{b}_2=a_1{k}_1{b}_1{k}_2=a_1{b}_1(b_1^{-1}{k}_1{b}_1)k_2\in (a_1{b}_1)N$ since $b_1^{-1}{k}_1{b}_1\in N$.
• Identity: $eN=N$. Inverses: $(a^{-1}N)(aN)=N$. Associativity: inherited from $G$.

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9. Quotient Groups

Strategy

1. $|G/N|=|G|/|N|$. If $=2$: $G/N\cong C_2$.
2. List cosets, or find homomorphism $\varphi$ with $\text{Ker}(\varphi )=N$ and apply FIT: $G/N\cong \text{Im}(\varphi )$.

$G$	$N$	$G/N$	Via
$\mathbb{Z}$	$n\mathbb{Z}$	${\mathbb{Z}}_n$	$\varphi (m)=[m{]}_n$
$S_n$	$A_n$	$C_2$	Signature
$S_3$	$C_3$	$C_2$	Index 2
${\text{GL}}_2$	${\text{SL}}_2$	${\mathbb{R}}^{\ast }$	$\text{det}$
${\text{GL}}_2$	$P$	$C_2$	$\text{sgn(det)}$
$P$	${\text{SL}}_2$	${\mathbb{R}}_{>0}$	$\text{det}$ at $P$