mth3003 final exam cheat sheet

MTH3003 Group Theory - Final Exam Cheat Sheet

Created by William Fayers

Good luck and have fun!! :)) - start at the §1 dispatcher, match your question, follow the arrows to a method, run the numbered steps, sanity-check (§14). Every recipe links to the definitions it needs. Straight from Simon’s notes (theorem numbers kept). Ez 100%.

§0. Notation, Groups, Key Facts

Notation & terms - every cross-reference bottoms out here.

Symbol / term	Plain meaning
$(G,\ast )$, $e_G$, $|G|$	group with operation $\ast$; identity $e_G$ = the element that does nothing in $G$ (e.g. $0$ in $(\mathbb{Z},+)$, $1$ in $({\mathbb{R}}^{\ast },\times )$); order = number of elements
$o(g)$	order of an element: least $n$ with $g^n=e$ (times to multiply $g$ to reach $e$)
abelian	$ab=ba$ - order of multiplication doesn’t matter
simple (Def 4.2.1)	only normal subgroups are $⟨e⟩$ and $G$
$H\le G$	subgroup: a smaller group inside $G$ (same operation)
$⟨g⟩$, $⟨S⟩$	generated: all you can build by multiplying; $⟨g⟩$ = powers of $g$, with $|⟨g⟩|=o(g)$; cyclic = whole group is some $⟨g⟩$
$gH$, $[G:H]$	coset = $H$ shifted by $g$, $\{gh:h\in H\}$; index = number of cosets $=|G|/|H|$
$N⊴G$	normal subgroup: $gN=Ng$ - the kind you can “divide by”
$G/N$	quotient group: the cosets of $N$, made a group by $(aN)(bN)=(ab)N$ (needs $N$ normal)
$\theta ,\phi ,\lambda$	homomorphism: $\theta (ab)=\theta (a)\theta (b)$ ($\lambda$ usually a group action)
$\ker \theta$, $\mathrm{Im}\theta$	kernel = elements sent to $e_H$; image = elements actually hit
$\cong$	isomorphic: a perfect 1-1 homomorphism (“same group, relabelled”)
conjugation, $C_G(x)$	replace $x$ by $g^{-1}xg$; centraliser = elements that commute with $x$
action, $Gx$, ${\mathrm{Stab}}_G(x)$, ${\mathrm{Fix}}_X(g)$	action = each $g$ permutes a set $X$; orbit = where $x$ can be sent (transitive = one orbit); stabiliser = the $g$‘s fixing $x$; fixed set = points fixed by one $g$
Sylow $p$-subgroup	biggest subgroup whose size is a power of the prime $p$
$G\times H$	direct product: pairs $(g,h)$ multiplied slot-by-slot
misc	$a\mid b$ ”$a$ divides $b$” ($\nmid$ = doesn’t); ${\mathbb{R}}^{\ast }$ nonzero reals; ${\mathbb{Z}}_n$ integers mod $n$; $(123)$ cycle

Axioms $(G,\ast )$: Closure, Associativity, Identity, Inverses.

Identities (Thm 2.1.11): $(g^{-1}{)}^{-1}=g$; cancel $g{h}_1=g{h}_2\Rightarrow h_1=h_2$; $(h_1{h}_2{)}^{-1}=h_2^{-1}{h}_1^{-1}$; $(g^{-1}hg{)}^n=g^{-1}{h}^{n}g$; identity/inverses unique.

Group	Order	Abelian?	Notes
$C_n=⟨(12\dots n)⟩$, ${\mathbb{Z}}_n$	$n$	yes	cyclic
$C_n\times C_m$ / any finite abelian	$nm$	yes	$\cong C_{nm}$ iff $\gcd (n,m)=1$ (§13)
$K_4\cong C_2\times C_2$	$4$	yes	$\{e,(12),(34),(12)(34)\}$; smallest non-cyclic
$(\mathbb{Z},+),({\mathbb{R}}^{\ast },\times )$	$\infty$	yes
$S_n$	$n!$	no ($n\ge 3$)	all perms of $\{1,\dots ,n\}$
$A_n$	$n!/2$	no ($n\ge 4$)	even perms; $A_n⊴S_n$; simple $n\ge 5$
$D_{2n}$	$2n$	no ($n\ge 3$)	$n$-gon symmetries ($D_6$ tri, $D_8$ square, $D_{10}$ pent)
${\mathrm{GL}}_2(\mathbb{R}),{\mathrm{SL}}_2(\mathbb{R})$	$\infty$	no	$\det \ne 0$; $\det =1$
$\mathrm{FS}(\mathbb{Z})$	$\infty$	no	finitary perms; $⊴\mathrm{Sym}(\mathbb{Z})$

§1. Dispatcher - Find Your Question

Q1-type (permutations / symmetry)

• Cycle notation, product, inverse, order, find element of order $d$, solve $\alpha \sigma \gamma =\beta$ → §2
• Odd/even (signature), is $g\in A_n$, list $A_n$ → §9
• $D_{2n}$ in cycles / simplify word / $⟨\rho ,\sigma ⟩=D_{2n}$ / symmetry group of a shape / “is it abelian” → §5; prove $S_3\cong D_6$ → §5+§8

Q2-type (structure: Lagrange, homs, iso theorems, simple)

• Prove $H\le G$ / group identity → §3; list $⟨g⟩$ / ”$G$ cyclic” / $|⟨b,c⟩|$ → §4
• State+prove Lagrange / rule out a subgroup order / cosets / Cauchy → §6
• Prove $N⊴G$ / $HN\le G$ / quotient group → §7
• Is it simple? / hom out of a simple group → §7
• Prove/disprove homomorphism / build an isomorphism → §8
• State+prove First Iso / identify a quotient via a hom / 3rd-theorem cancellation → §9 (theorems) + §8 (build map)

Q3/Q4-type (actions / Sylow / abelian)

• Is $\lambda$ an action / verify-or-refute / Cayley → §10
• Orbit & stabiliser / Orbit-Stabiliser / rule out orbit size / orbit counting → §11
• State Sylow / Sylow example / count $a(p)$ / “not simple” / subgroup of order $p^r$ → §12
• Classify abelian groups of order $n$ / subgroups of abelian / element of order $d$ / $G\cong M\times N$ / $C_n\times C_m$ abelian → §13

Each question: part (a) ~9 marks (often “state & prove”), (b)~8, (c)~8 (stretch). Answer best 3 of 4.

§2. Permutations

Cycle $(a_1\dots a_r)$: $a_i\mapsto a_{i+1}$, $a_r\mapsto a_1$, rest fixed. Disjoint cycles commute (1.2.7); every $g$ = unique product of disjoint cycles (1.3.4). Products act right-to-left.

• Two-row → cycles: start at smallest unused $x$, follow $x\to g(x)\to \dots$ till back to $x$ (closes a cycle), repeat on unused; drop fixed points.
• Cycles → two-row: read off $a_i\mapsto a_{i+1}$; unlisted points fixed.
• Product $gh$: chase each point right-to-left through every factor, close each cycle, repeat. (Disjoint factors can be done independently.) ex $(123)(24)$ (do $(24)$ first): $1\to 2$, $2\to 4$, $4\to 3$, $3\to 1$ → $(1243)$.
• Inverse: reverse each cycle, $(a_1\dots a_r{)}^{-1}=(a_r\dots a_1)$; invert a product factor-by-factor.
• Order: $o(g)=\mathrm{lcm}$ of cycle lengths.
• Find element of order $d$ in $S_n$: choose disjoint cycle lengths summing to $\le n$ with $\mathrm{lcm}=d$ (e.g. $20=\mathrm{lcm}(4,5)$, lengths $4+5\le n$). None exists if it would need a cycle longer than $n$ (e.g. order $19$ in $S_{10}$: needs a $19$-cycle).
• Solve $\alpha \sigma \gamma =\beta$: cancel outward - $\sigma ={\alpha }^{-1}\beta {\gamma }^{-1}$ (left factor inverts on left, right factor on right). Then compute and verify by substituting back. ex (WP1.4): $(132)\sigma =(12)(34)$ → $\sigma =(132{)}^{-1}(12)(34)=(123)(12)(34)=(134)$.
• Odd/even → §9.

ex (Mock A1a): $g=(123)(45)$, $o(g)=\mathrm{lcm}(3,2)=6$; $⟨g⟩=\{e,g,(132),(45),(123),(132)(45)\}$.

§3. Subgroups

• Prove $H\le G$ (Quick Subgroup Test, Thm 2.2.5): show 1) $e\in H$, 2) $h_1,h_2\in H\Rightarrow h_1{h}_2\in H$, 3) $h\in H\Rightarrow h^{-1}\in H$ (or combine 2-3: $h_1{h}_2^{-1}\in H$). Standard targets: $H\cap K$ (use both), $\mathrm{Im}\theta$ (§8), ${\mathrm{Stab}}_G(x)$ (§11), $A_n$ (§9), $\mathrm{FS}(X)$ (support stays finite, §7).
• Prove a group identity: expand and cancel $g{g}^{-1}=e$. Key ones: $(g^{-1}hg{)}^n=g^{-1}{h}^{n}g$; $(h_1{h}_2{)}^{-1}=h_2^{-1}{h}_1^{-1}$.
• Order facts: $|H|=1⟺H=⟨e⟩$; finite $|H|=|G|⟺H=G$ (fails if infinite: $\mathbb{Z}<\mathbb{Q}$).

§4. Cyclic Groups

$⟨g⟩=\{g^n\}$, $|⟨g⟩|=o(g)$.

• List $⟨g⟩$: compute $e,g,g^2,\dots$ until $e$ repeats ($o(g)$ elements). $⟨g^k⟩$ likewise.
• Prove $|⟨g⟩|=o(g)$: division algorithm $m=qo(g)+r$ $\Rightarrow g^m=g^r$, so exactly $g^0,\dots ,g^{o(g)-1}$.
• Prove $G$ cyclic: if $|G|=p$ prime, any $g\ne e$ has $|⟨g⟩|\mid p$ (Lagrange §6) so $=p$, thus $⟨g⟩=G$. For ${\mathbb{Z}}_n$: $[m{]}_n=([1{]}_n{)}^m$ so ${\mathbb{Z}}_n=⟨[1{]}_n⟩$.
• Disprove "$|⟨b,c⟩|=o(b)o(c)$": counterexample $b=(123)(45),c=(45)$: $⟨b,c⟩=⟨b⟩$, order $6\ne 12$.

§5. Dihedral & Symmetry Groups

$D_{2n}=\{e,\rho ,\dots ,{\rho }^{n-1},\sigma ,\sigma \rho ,\dots ,\sigma {\rho }^{n-1}\}$, order $2n$; ${\rho }^n={\sigma }^2=e$, $\boxed{\rho \sigma =\sigma {\rho }^{-1}}$ (so ${\rho }^k\sigma =\sigma {\rho }^{-k}$). $\rho =(12\dots n)$, $\sigma$ = reflection through vertex $1$.

• Write $D_{2n}$ in cycle notation: rotations ${\rho }^k=$ “shift by $k$”; reflections - draw the $n$-gon and read off, or compute $\sigma {\rho }^i$.
• Prove $\rho \sigma =\sigma {\rho }^{-1}$: $\rho \sigma$ is a reflection so $o(\rho \sigma )=2$, i.e. $(\rho \sigma {)}^2=e$; rearrange $\rho \sigma =(\rho \sigma {)}^{-1}={\sigma }^{-1}{\rho }^{-1}=\sigma {\rho }^{-1}$.
• Simplify a word to ${\rho }^k$ or $\sigma {\rho }^k$: push every $\sigma$ to the left via $\rho \sigma =\sigma {\rho }^{-1}$, then reduce mod ${\rho }^n=e$, ${\sigma }^2=e$.
• Prove $⟨\rho ,\sigma ⟩=D_{2n}$: every word reduces (above) to ${\sigma }^k{\rho }^j$, $k\in \{0,1\}$, $j\in \{0,\dots ,n-1\}$ = the $2n$ listed elements.
• Symmetry group of a labelled shape: write each rotation and reflection as a permutation of the labels; collect them.
• Construct a non-$n$-gon shape with $D_{2n}$ symmetry: regular $n$-gon + an identical decoration on each edge (e.g. hexagon + square per edge $\to D_{12}$).
• Is the symmetry group abelian? Identify it by counting symmetries: $m$ rotations and no reflections → $\cong C_m$ (cyclic, abelian); $m$ rotations and $m$ reflections → $\cong D_{2m}$ (non-abelian for $m\ge 3$). ex (Mock B1b): a pinwheel (4-fold rotation, no reflections) → $C_4$ → abelian. (Which groups are abelian: §0.)
• Prove $S_3\cong D_6$: build the iso (§8) on generators $\rho \mapsto (123),\sigma \mapsto (12)$; check relations hold and images are all 6 of $S_3$.

§6. Cosets & Lagrange

Coset $gH=\{gh:h\in H\}$ (Def 4.1.1); $[G:H]$ = distinct cosets. $aH=bH⟺a^{-1}b\in H$ (4.1.3); cosets are disjoint-or-equal, all size $|H|$ (4.1.4).

• List cosets / find index: start $eH=H$; pick $g\notin$ listed, form $gH$; repeat until $G$ used up; $[G:H]$ = count. ex $S_3$, $H=⟨(12)⟩=\{e,(12)\}$: $H=\{e,(12)\}$, $(123)H=\{(123),(13)\}$, $(132)H=\{(132),(23)\}$ → $[S_3:H]=3$.
• Lagrange (Thm 4.2.5) - “state & prove”:
  • Statement: if $H\le G$ with $G$ finite, then $|G|=[G:H]|H|$; in particular $|H|\mid |G|$.
  • Terms: $H\le G$ = subgroup; $|G|,|H|$ = orders (number of elements); coset $gH=\{gh:h\in H\}$; index $[G:H]$ = number of distinct cosets of $H$ in $G$.
  • Proof: every $g\in G$ lies in a coset (4.1.4), so $G=g_{1}H\cup \cdots \cup g_{n}H$ with the $n=[G:H]$ distinct cosets; distinct cosets are disjoint (4.1.4) and all have size $|H|$ (bijection $h\mapsto g_{i}h$); hence $|G|=|g_{1}H|+\cdots +|g_{n}H|=n|H|=[G:H]|H|$, so $|H|\mid |G|$. $■$
• Rule out a subgroup/element of order $k$: if $k\nmid |G|$, impossible. ($o(g)\mid |G|$ via $⟨g⟩$, §4.) Converse is false ($A_4$, no order-6).
• Cauchy (4.2.6): prime $p\mid |G|\Rightarrow$ an element of order $p$ exists.

§7. Normal Subgroups, Quotients & Simple Groups

$N⊴G$ (Def 4.1.5): $g^{-1}kg\in N$ $\forall k\in N,g\in G$; equiv. $gN=Ng$ $⟺g^{-1}Ng=N$.

• Prove $N⊴G$ - pick the cheapest route:
  1. Direct: fix $g\in G,k\in N$, show $g^{-1}kg\in N$ (do it for all $g,k$).
  2. Abelian $G$ ⇒ automatic (which are abelian: §0).
  3. $N=\ker \theta$ for some hom $\theta$ ⇒ normal (§8) - the slickest; e.g. $A_n=\ker \sigma$, ${\mathrm{SL}}_2=\ker \det$.
  4. Index 2 ($[G:N]=2$, §6) ⇒ normal.
  5. Invariant under conjugation by a formula: $\det (g^{-1}Ag)=\det A$, so ${\mathrm{SL}}_2=\{\det =1\}$ and $P=\{A:\det A>0\}$ are normal in ${\mathrm{GL}}_2$; and $|\mathrm{supp}(g^{-1}hg)|=|\mathrm{supp}(h)|$ (support = points moved), so $\mathrm{FS}(X)⊴\mathrm{Sym}(X)$.
  6. Unique Sylow $p$-subgroup ⇒ normal (§12).
• Order Switching (Lem 4.2.3) $N⊴G$: $gk=k^{\prime }g$, $kg=g{k}^{\prime \prime }$ ($k^{\prime },k^{\prime \prime }\in N$). Key Lemma (4.2.4): $H\le G,N⊴G\Rightarrow HN=NH$, $H\cap N⊴H$, $HN\le G$. (Use these to prove $HN\le G$ via Quick Subgroup Test, swapping orders with Order Switching.)
• Quotient $G/N$ (needs $N⊴G$): elements $gN$, $(aN)(bN)=(ab)N$, $|G/N|=|G|/|N|$. Identify it: find a surjective hom with kernel $N$ and apply FIT (§9) - e.g. $|G/N|=2\Rightarrow C_2$.
• Is $G$ simple? (only normal subgroups $⟨e⟩,G$):
  • $C_p$ ($p$ prime) yes (subgroup orders $1,p$ only).
  • No if abelian with a proper nontrivial subgroup (e.g. $K_4$ - all subgroups normal).
  • $S_n$ ($n\ge 3$) no ($A_n⊴S_n$, §9).
  • $M\times N$ no ($⟨e⟩\times N⊴M\times N$, §13). E.g. $C_5\times C_9$.
  • General order: use Sylow to force a unique (normal) Sylow ⇒ not simple (§12).
• Hom out of a simple group (Mock B2c, worked):
  • Claim: if $G$ is simple and $\phi :G\to H$ is a homomorphism, then either $\phi (g)=e_H$ for all $g$ (trivial), or $\phi$ is injective.
  • Proof: $\ker \phi ⊴G$ (§8). Since $G$ is simple, its only normal subgroups are $⟨e⟩$ and $G$, so $\ker \phi =⟨e⟩$ or $\ker \phi =G$. If $\ker \phi =G$, every $g$ maps to $e_H$ (trivial); if $\ker \phi =⟨e⟩$, then $\phi$ is injective (§8). $■$

§8. Homomorphisms & Isomorphisms

Hom: $\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)$. Props (5.2.4): $\theta (e_G)=e_H$, $\theta (g^{-1})=\theta (g{)}^{-1}$, $\theta (g^m)=\theta (g{)}^m$, $\mathrm{Im}\theta \le H$, $\ker \theta ⊴G$.

• Prove a map is a hom: check $\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)$ for general $g_1,g_2$. Disprove: one counterexample. ($\varphi (g)=g^k$ is a hom $⟺G$ abelian, since $(g_1{g}_2{)}^k=g_1^k{g}_2^k⟺$ commute.)
• Prove $\theta (e)=e$: $\theta (g)=\theta (eg)=\theta (e)\theta (g)$, cancel. $\theta (g^m)=\theta (g{)}^m$: induction. $\ker \theta ⊴G$: $\theta (g^{-1}kg)=\theta (g{)}^{-1}{e}_H\theta (g)=e_H$. $\mathrm{Im}\theta \le H$: Quick Subgroup Test (§3).
• Prove an isomorphism $G\cong H$: build $\theta$ (define on generators, extend), then check 1) hom, 2) injective ($\ker \theta =⟨e⟩$), 3) onto. Shortcut: if $|G|=|H|$ finite, injective $⟺$ onto (check one). ex $S_3\cong D_6$ (here $D_6=⟨\rho ,\sigma ⟩$ with $\rho$ = a rotation and $\sigma$ = a reflection, §5): set $\theta (\rho )=(123),\theta (\sigma )=(12)$; the defining relations carry over ($(123{)}^3=e$, $(12{)}^2=e$, $(123)(12)=(12)(123{)}^{-1}$) so $\theta$ is a hom; its 6 images are all of $S_3$, so (equal orders) it’s a bijection ⇒ iso.
• Power map $g\mapsto g^k$ on $C_n$ is an iso $⟺\gcd (k,n)=1$.
• $\ker \theta$ normal is the engine for §7 route 3 and §9.

§9. Isomorphism Theorems & the Alternating Group

• First Isomorphism Theorem (Thm 6.0.1) - “state & prove”:
  • Statement: if $\theta :G\to H$ is a homomorphism with kernel $K$ and image $I$, then $G/K\cong I$ - via the map ${\theta }^{\ast }(gK)=\theta (g)$.
  • Terms: homomorphism $\theta (ab)=\theta (a)\theta (b)$; kernel $K=\{g:\theta (g)=e_H\}$; image $I=\{\theta (g):g\in G\}$; quotient $G/K$ = cosets of $K$ with $(aK)(bK)=(ab)K$; $\cong$ = isomorphic (a bijective homomorphism).
  • Proof (4 checks): well-defined $g_{1}K=g_{2}K\Rightarrow g_1^{-1}{g}_2\in K\Rightarrow \theta (g_1^{-1}{g}_2)=e_H\Rightarrow \theta (g_1)=\theta (g_2)$; hom ${\theta }^{\ast }(g_{1}K{g}_{2}K)={\theta }^{\ast }(g_1{g}_{2}K)=\theta (g_1{g}_2)=\theta (g_1)\theta (g_2)$; onto each element of $I$ is $\theta (g)={\theta }^{\ast }(gK)$; 1-1 ${\theta }^{\ast }(g_{1}K)={\theta }^{\ast }(g_{2}K)\Rightarrow \theta (g_1)=\theta (g_2)\Rightarrow \theta (g_1^{-1}{g}_2)=e_H\Rightarrow g_1^{-1}{g}_2\in K\Rightarrow g_{1}K=g_{2}K$. $■$
• Use FIT to identify a quotient $G/N$: spot a surjective hom $G\to H$ with $\ker =N$; then $G/N\cong H$. ex ${\mathrm{GL}}_2(\mathbb{R})/{\mathrm{SL}}_2(\mathbb{R})$: $\det :{\mathrm{GL}}_2\to {\mathbb{R}}^{\ast }$ is a hom, onto, with $\ker (\det )=\{A:\det A=1\}={\mathrm{SL}}_2$; FIT ⇒ ${\mathrm{GL}}_2/{\mathrm{SL}}_2\cong {\mathbb{R}}^{\ast }$. Table: $\mathbb{Z}/n\mathbb{Z}\cong {\mathbb{Z}}_n$ ($m\mapsto [m{]}_n$); $S_n/A_n\cong C_2$ (signature); ${\mathrm{GL}}_2/{\mathrm{SL}}_2\cong {\mathbb{R}}^{\ast }$ ($\det$); $⟨g⟩/⟨g^d⟩\cong C_d$.
• Second (6.0.4): $H/(H\cap N)\cong (HN)/N$ (map $h\mapsto hN$, kernel $H\cap N$).
• Third (6.0.5): $N\le M$, both $⊴G\Rightarrow (G/N)/(M/N)\cong G/M$ (“cancellation”; map $gN\mapsto gM$).

Signature $\sigma :S_n\to \{1,-1\}$ (Def 7.2.1, via $\Delta ={\prod }_{i<j}(x_i-x_j)$). $A_n=\ker \sigma$ (Def 7.3.1). Cycle shape $(r_1\ge \cdots \ge r_m)$ (7.1.1).

• Compute $\sigma (g)$: $\sigma (g)=\prod (-1{)}^{r_i-1}$ over disjoint cycles (Cor 7.2.3). Quick: count even-length cycles - odd count ⇒ odd. $\sigma ((1\dots n))=(-1{)}^{n-1}$ (even iff $n$ odd). ex (Mock B1a): $(4315)(267)$ = 4-cycle (even length → sign $-1$) · 3-cycle (sign $+1$) → $\sigma =-1$, odd. For a product $gh$, just multiply signs: $\sigma (gh)=\sigma (g)\sigma (h)$.
• $\sigma$ is a homomorphism (Prop 7.2.2, examinable). Statement: $\sigma (gh)=\sigma (g)\sigma (h)$. Proof: $(gh)\Delta =g(h\Delta )=g(\sigma (h)\Delta )=\sigma (h)(g\Delta )=\sigma (h)\sigma (g)\Delta$, so $\sigma (gh)=\sigma (g)\sigma (h)$. $■$ Also $\sigma (g^{-1})=\sigma (g)$ (Prop 7.2.5), $\sigma (g^{-1}hg)=\sigma (h)$; for a product, multiply signs (write as disjoint cycles first if computing directly).
• Is $g\in A_n$? $⟺\sigma (g)=1$.
• $A_n⊴S_n$ and $|A_n|=n!/2$ - “state & prove”:
  • Statement: the even permutations $A_n=\ker \sigma$ form a normal subgroup of $S_n$, with $|A_n|=n!/2$.
  • Proof: subgroup (Quick Subgroup Test): $\sigma (e)=1$, $\sigma (gh)=\sigma (g)\sigma (h)=1$, $\sigma (g^{-1})=\sigma (g)=1$. normal (Thm 7.3.5): $\sigma (g^{-1}hg)=\sigma (g^{-1})\sigma (h)\sigma (g)=\sigma (g^{-1})\sigma (g)=\sigma (g^{-1}g)=\sigma (e)=1$ (as $\sigma (h)=1$), so $g^{-1}hg\in A_n$. order (Thm 7.3.2): $\sigma$ is onto $\{1,-1\}$ with $\ker \sigma =A_n$, so FIT gives $S_n/A_n\cong \{1,-1\}$, whence $|A_n|=|S_n|/2=n!/2$. $■$
  • Simple for $n\ge 5$ (proof not examinable); $A_4$ is not simple ($K_4⊴A_4$).
• List $A_n$ via cycle shapes (shape: sign, count): $S_3$ $\varnothing (+,1),(2)(-,3),(3)(+,2)\Rightarrow A_3=\{e,(123),(132)\}$. $S_4$ even: $\varnothing ,(2,2)(3),(3)(8)\Rightarrow |A_4|=12$. $S_5$ even: $\varnothing (1),(2,2)(15),(3)(20),(5)(24)\Rightarrow |A_5|=60$ [$(2,2)=5\cdot 3$, $(3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2$, $(5)=5!/5$].

§10. Group Actions & Cayley

Action (Def 8.1.1) $\lambda :G\times X\to X$: (1) $\lambda (e_G)x=x$ ($e_G$ = the identity of the group $G$), (2) $\lambda (fg)x=\lambda (f)(\lambda (g)x)$. Same thing as a hom $\lambda :G\to \mathrm{Sym}(X)$ (8.1.4). $\ker \lambda ⊴G$.

• Verify a map is an action: check 0) closure $\lambda (g)x\in X$, 1) identity $\lambda (e_G)x=x$ (use $G$‘s own identity $e_G$ - e.g. $0$ in $(\mathbb{Z},+)$, not $1$), 2) compatibility $\lambda (fg)x=\lambda (f)(\lambda (g)x)$.
• Show it’s NOT an action: exhibit the failing axiom. Closure (WP8.4: $(\mathbb{Z},+)$ on $\mathbb{N}$, $\lambda (-5)1=-4\notin \mathbb{N}$); identity (Mock A3a: $(\mathbb{Z},+)$ on $\mathbb{Z}$ by $mn$ - here $e_G=0$, and $\lambda (e_G)n=0\cdot n=0\ne n$); compatibility (8.1.10: $\lambda (g)x=g^{-1}xg$).
• A hom $G\to \mathrm{Sym}(X)$ is automatically an action (and vice versa) - the two are the same data (WP8.8).
• Standard actions on $X=G$: regular $\lambda (g)x=gx$; (left) conjugation $\lambda (g)x=gx{g}^{-1}$. On subsets: $\lambda (g)\{y_1,\dots \}=\{g{y}_1,\dots \}$ (check it lands in the right-sized subsets).
• Cayley: every $G\cong$ a subgroup of $\mathrm{Sym}(G)$. Proof: regular action has $\ker =⟨e⟩$ ($gx=x\Rightarrow g=e$), so FIT (§9) gives $G\cong \mathrm{Im}\lambda$.

§11. Orbits, Stabilisers, Counting

Orbit $Gx=\{\lambda (g)x:g\in G\}$ (8.3.1); transitive if $Gx=X$. Stabiliser ${\mathrm{Stab}}_G(x)=\{g:\lambda (g)x=x\}\le G$ (8.3.2). Orbits partition $X$ (9.1.0). Fixed set ${\mathrm{Fix}}_X(g)=\{x:\lambda (g)x=x\}$ (9.2.1).

• Compute orbit & stabiliser of $x$: work out $\lambda (g)x$ for every $g\in G$, then:
  • orbit = the set of distinct images $\{\lambda (g)x\}$ - a subset of $X$ (the places $x$ lands);
  • stabiliser = the set of those $g$ that give back $x$ $\{g:\lambda (g)x=x\}$ - a subset of $G$ (the group elements that fix $x$).
• Deduce if the action is transitive (from the orbit): transitive $⟺$ the orbit is all of $X$, i.e. $Gx=X$ (equivalently $|Gx|=|X|$). So compute the orbit, then compare its size to $|X|$: equal → transitive; smaller → not transitive.
• Prove ${\mathrm{Stab}}_G(x)\le G$: Quick Subgroup Test (§3): $\lambda (e_G)x=x$; $g,h$ fix $x\Rightarrow gh$ fixes $x$; $g$ fixes $x\Rightarrow g^{-1}$ does.
• Orbit-Stabiliser (Thm 9.1.1) - “state & prove”:
  • Statement: for all $x\in X$, $|Gx|=[G:{\mathrm{Stab}}_G(x)]=|G|/|{\mathrm{Stab}}_G(x)|$ (finite $G$).
  • Terms: orbit $Gx=\{\lambda (g)x:g\in G\}$; stabiliser ${\mathrm{Stab}}_G(x)=\{g\in G:\lambda (g)x=x\}$; $[G:{\mathrm{Stab}}_G(x)]$ = index (number of cosets).
  • Proof: the map $gS\mapsto \lambda (g)x$ ($S={\mathrm{Stab}}_G(x)$) is a bijection $\{$cosets of $S\}\to Gx$, since $\lambda (g)x=\lambda (h)x⟺h^{-1}g\in S⟺gS=hS$ (coset rule §6); so $|Gx|=[G:S]$, and Lagrange gives $|G|/|S|$. $■$
• Find an orbit/stabiliser size when you know the other: divide $|G|$.
• Rule out an orbit of size $k$ / a transitive action on $|X|$: if $k\nmid |G|$ (resp. $|X|\nmid |G|$), impossible (Orbit-Stab forces integer). E.g. no orbit of size 4 in $S_3$ ($4\nmid 6$); $D_{10}$ not transitive on a triangle ($3\nmid 10$).
• Conjugation action: ${\mathrm{Stab}}_G(x)=C_G(x)$ = centraliser (things commuting with $x$); orbit = conjugacy orbit. ex (Mock A3): in $S_3$, $\mathrm{Stab}((12))=\{e,(12)\}$, orbit $=\{(12),(13),(23)\}$ (not transitive).
• Stabiliser of a subset (Mock B4): ${\mathrm{Stab}}_{S_5}(\{1,2,3\})=S_{\{1,2,3\}}\times S_{\{4,5\}}$, order $3!2!=12$; orbit = all $3$-subsets (size 10); $10\cdot 12=120$. ✓
• ${\bigcap }_x{\mathrm{Stab}}_G(x)=\ker \lambda ⊴G$ (Mock B4c): quick - it’s the kernel of the action-hom (§10), and any kernel is normal (§8). Direct - $h\in N$ fixes every point, so for any $x$: $\lambda (g^{-1}hg)x=\lambda (g^{-1})\lambda (h)\lambda (g)x=\lambda (g^{-1})\lambda (g)x=x$; thus $g^{-1}hg\in N$, so $N⊴G$.
• Count orbits (Orbit Counting / Burnside, 9.2.3): $\#\text{orbits}=\frac{1}{|G|}{\sum }_g|{\mathrm{Fix}}_X(g)|$. Recipe for colourings with $c$ colours: each $g$ fixes $c^{\#\text{cycles of }g\text{ on the slots}}$; tabulate, sum, divide by $|G|$. Octagon edges, 3 col, $D_{16}$: $498$. Cube faces, 3 col, rotations (24, no reflections): $57$. [Proof: double-count $P=\{(x,g):gx=x\}$: by $g$, $|P|={\sum }_g|{\mathrm{Fix}}_X(g)|$; by $x$, $|P|={\sum }_x|{\mathrm{Stab}}_G(x)|={\sum }_x|G|/|Gx|=m|G|$ (orbits partition $X$, each contributes $|G|$). Equate $\Rightarrow m=\frac{1}{|G|}{\sum }_g|{\mathrm{Fix}}_X(g)|$.]

§12. Sylow’s Theorems

$|G|=p^{r}t$ with $p\nmid t$ ($t$ = the part of $|G|$ coprime to $p$). $p$-group (10.1.1): a group of order $p^r$. Sylow $p$-subgroup (10.1.3): a subgroup of order $p^r$ (the largest power of $p$ dividing $|G|$). $a(p)$ = the number of Sylow $p$-subgroups.

• State Sylow (define every term, Mock A4a):
  1. First (10.1.5): $p\mid |G|\Rightarrow$ a Sylow $p$-subgroup exists. (Generalises Cauchy §6.)
  2. Second (10.1.6): every $p$-subgroup lies in a conjugate of a Sylow $p$-subgroup; all Sylow $p$-subgroups conjugate.
  3. Third (10.1.7): $a(p)\mid t$ and $a(p)\equiv 1(\mathrm{mod}p)$. (Proofs not examinable.)
• Give a Sylow example: factor $|G|$ for the order $p^r$, then write down any subgroup of that order built from elements of $p$-power order. Its type depends on $p^r$ - not always dihedral:
  • order $p$ ($r=1$): cyclic $C_p=⟨p\text{-cycle}⟩$ - e.g. Sylow 5 of $S_5$ is $⟨(12345)⟩\cong C_5$.
  • order $p^2$: $C_{p^2}$ or $C_p\times C_p$ - e.g. Sylow 3 of $S_6$ ($720=2^4{3}^{2}5$) is $⟨(123)⟩\times ⟨(456)⟩\cong C_3\times C_3$ (abelian, not dihedral).
  • order 8 ($2^3$, in $S_4,S_5$): the square’s symmetries $D_8=⟨(1234),(13)⟩$ - this is the dihedral case (§5).
  • (only need it to exist? don’t construct - just cite Sylow First, next bullet.)
• Prove a subgroup of order $p^r$ exists: factor $|G|$, quote Sylow First.
• Count / bound $a(p)$ (via Sylow Third): write $|G|=p^{r}t$ with $p\nmid t$ (so $t$ = the part of $|G|$ coprime to $p$, and $a(p)$ = number of Sylow $p$-subgroups). Then $a(p)\mid t$ and $a(p)\equiv 1(\mathrm{mod}p)$. So list the divisors of $t$, keep those $\equiv 1(\mathrm{mod}p)$ - that’s the candidate set. To show $a(p)\ge k$: exhibit distinct Sylow subgroups (e.g. two squares on different 4-point sets ⇒ $a(2)\ge 2$, so $\ge 3$ if the candidates are $\{1,3,\dots \}$).
• Prove ”$G$ is not simple” (i.e. find a normal subgroup that is neither $⟨e⟩$ nor $G$):
  1. Why this works: simple = the only normal subgroups are $⟨e⟩$ and $G$. So you just need one proper, nontrivial normal subgroup. Best source: a Sylow $p$-subgroup that is unique, because $a(p)=1\Rightarrow$ it’s normal (§7).
  2. Factorise $|G|=p^{r}t$ and pick a prime $p$ dividing it - try the largest prime first (most likely to force $a(p)=1$).
  3. Pin down $a(p)$ with Sylow Third: write the divisors of $t$, then keep only those $\equiv 1(\mathrm{mod}p)$.
  4. If only $a(p)=1$ survives: the Sylow $p$-subgroup $P$ is unique, hence $P⊴G$; and $|P|=p^r$ lies strictly between $1$ and $|G|$, so $P$ is proper and nontrivial ⇒ $G$ is not simple. (If $a(p)$ is not forced to 1, try another prime.)
  • Worked example, $|G|=63$: factor $63=3^2\cdot 7$. Take $p=7$ (largest): then $t=9$, divisors $1,3,9$; their remainders mod 7 are $1,3,2$, so only $a(7)=1$ survives. So the Sylow 7-subgroup $P$ ($|P|=7$) is unique ⇒ $P⊴G$; since $1<7<63$, $P$ is proper and nontrivial ⇒ $G$ not simple. $■$
  • (Why not $p=3$? Then $t=7$, divisors $1,7$, and both are $\equiv 1(\mathrm{mod}3)$, so $a(3)\in \{1,7\}$ - not forced to 1, so it proves nothing. That’s why we picked the larger prime 7.)
• Classify small order: unique normal Sylows $H,K$ with $H\cap K=⟨e⟩$, $|HK|=|G|$ ⇒ $G\cong H\times K$ (§13). E.g. order 15 ⇒ $C_{15}$.

§13. Direct Products & Finite Abelian Groups

External $G\times H$: $(g_1,h_1)(g_2,h_2)=(g_1{g}_2,h_1{h}_2)$; $|G\times H|=|G||H|$; $G\times H\cong H\times G$ (11.1.1).

• Prove $G\times H\cong H\times G$: $\theta (g,h)=(h,g)$; check hom + bijection (§8).
• Prove $K\times J\le G\times H$ ($K\le G,J\le H$): Quick Subgroup Test componentwise (§3).
• Internal direct product (Thm 11.1.3, examinable) - “state & prove”:
  • Statement: if $M,N⊴G$ with $M\cap N=⟨e⟩$, then $MN\le G$ (Key Lemma §7) and $MN\cong M\times N$.
  • Terms: $M,N⊴G$ = normal subgroups (§7); $⟨e⟩$ = trivial subgroup $\{e\}$; $MN=\{mn:m\in M,n\in N\}$; $M\times N$ = external direct product (above).
  • Proof: (commuting, Lemma 11.1.2) $n^{-1}mn\in M$ ($M$ normal) so $n^{-1}mn{m}^{-1}\in M$; $mn{m}^{-1}\in N$ ($N$ normal) so $n^{-1}mn{m}^{-1}\in N$; thus $\in M\cap N=⟨e⟩$, giving $mn=nm$. Then each $g\in MN$ is uniquely $mn$ (if $mn=m_1{n}_1$ then $m^{-1}{m}_1=n{n}_1^{-1}\in M\cap N=⟨e⟩$), so $\theta (mn)=(m,n)$ is well-defined, a homomorphism (by commuting), 1-1 and onto. $■$
  • To detect $G\cong M\times N$: find normal $M,N$ with trivial intersection and $|MN|=|G|$.
• Prove $C_n\times C_m$ abelian: $(a,b)(c,d)=(ac,bd)=(ca,db)=(c,d)(a,b)$ (each factor abelian).
• $C_n\times C_m\cong C_{nm}⟺\gcd (n,m)=1$ (11.2.1/2). So $C_3\times C_2\cong C_6$.
• Fundamental Theorem (11.3.2): every finite abelian $G\cong C_{p_1^{e_1}}\times \cdots \times C_{p_k^{e_k}}$ (prime-power cyclic factors), unique up to order. [Fin. gen. (11.3.3): ${\mathbb{Z}}^{e_0}\times$torsion, $e_0=$ rank.]
• Classify all abelian groups of order $n$ (FTfAG): factor $n=\prod p_i^{a_i}$; for each prime, list the partitions of the exponent $a_i$ (ways to write $a_i$ as a sum of positive integers, order ignored); a partition $a_i=b_1+b_2+\dots$ gives factors $C_{p^{b_1}}\times C_{p^{b_2}}\times \cdots$; combine across primes. ($\#$ abelian groups of order $p^a$ = partitions of $a$.) ex (Mock B3a) $27=3^3$: partitions of 3 are $(3),(2,1),(1,1,1)$ → $C_{27}$, $C_9\times C_3$, $C_3\times C_3\times C_3$ - three groups.
• Find all subgroups of an abelian group (up to isomorphism) - three steps:
  • (1) Possible orders: by Lagrange (§6), each subgroup’s order divides $|G|$.
  • (2) Exhibit one subgroup of each of those orders.
  • (3) Identify up to isomorphism - the step that completes the list (don’t skip): show each order has only one possible type, so any subgroup of that order matches one you found. Tools: order 1 = trivial group; order $|G|$ = $G$ itself; prime order $\Rightarrow$ cyclic $\cong C_p$ (§4).
  • All subgroups here are automatically normal (since $G$ is abelian, §7).
  • Example - $C_3\times C_2$, order 6 (write $C_3=⟨a⟩$, $C_2=⟨b⟩$):
    • Orders dividing 6: $1,2,3,6$.
    • Exhibit: $⟨e⟩$; $⟨(e,b)⟩\cong C_2$; $⟨(a,e)⟩\cong C_3$; $G\cong C_3\times C_2$.
    • Iso step: any order-2 subgroup $\cong C_2$, any order-3 $\cong C_3$ (prime ⇒ cyclic); order 1 = trivial; order 6 $=G$.
    • Four up to isomorphism: trivial, $C_2$, $C_3$, $C_3\times C_2(\cong C_6)$.
• Count abelian groups of order $n$ with an element of order $d$:
  1. Key fact: the order of an element $(g_1,g_2,\dots )$ of a direct product = lcm of the orders of its components. So an element of order $d$ exists iff, prime by prime, the factors can supply each prime-power in $d$.
  2. For each prime power $p^k$ exactly dividing $d$: you need a cyclic factor $C_{p^j}$ with $j\ge k$ in the $p$-part (only such a factor holds an element of order $p^k$). Primes not in $d$ are unconstrained.
  3. So count, per relevant prime, the partitions of its exponent that include a part $\ge k$; multiply across primes (free primes contribute all their partitions).
  • Worked: order $540=2^2\cdot 3^3\cdot 5$, want an element of order $45=3^2\cdot 5$ (so need order $9$ from the 3-part and order $5$ from the 5-part):
    • 3-part $3^3$ (partitions of 3): $C_{27}$ ✓, $C_9\times C_3$ ✓ (each has a factor $\ge C_9$), $C_3\times C_3\times C_3$ ✗ (max element order $3$). → 2 of 3.
    • 5-part $5^1$: only $C_5$ - has an element of order 5 ✓. → 1.
    • 2-part $2^2$: irrelevant to order 45, so free: $C_4$ or $C_2\times C_2$. → 2.
    • Total $2\times 1\times 2=4$ (out of the $3\times 1\times 2=6$ abelian groups of order 540).
• $C_{p^s}\lesssim C_{p^r}$ for $s\le r$ (take ${\tau }^{p^{r-s}}$); hence abelian converse to Lagrange: abelian $G$, $k\mid |G|\Rightarrow G$ has a subgroup of order $k$ (build it factor-by-factor via FTfAG).

§14. Pitfalls & Sanity Checks

Pitfalls

• Permutation products act right-to-left.
• Converse of Lagrange is false ($A_4$).
• $\varphi (g)=g^k$ is a hom only if $G$ abelian.
• Quotient $G/N$ needs $N$ normal - check first.
• Action checks: don’t forget closure $\lambda (g)x\in X$, then identity, then compatibility.
• Sylow $a(p)$ needs both $a(p)\mid t$ and $a(p)\equiv 1(\mathrm{mod}p)$.
• Cube colouring = rotations only (24); octagon (2D) allows reflections (16).
• FTfAG factors are prime powers; element order = lcm of component orders.

Sanity checks

• Orders divide: $o(g)\mid |G|$, $|H|\mid |G|$, class/orbit sizes $\mid |G|$.
• $|G/N|=|G|/|N|$ and $|Gx|\cdot |{\mathrm{Stab}}_G(x)|=|G|$ are integers.
• Isomorphism check: $|G|=|H|$, and it preserves order/abelian/cyclic.
• Abelian-group count = product of partition-counts of the prime exponents.
• After solving $\alpha \sigma \gamma =\beta$ or building an iso: substitute back.