If you are looking to build your own "Overleaf" document, here is the code for a high-quality solution set covering selected exercises (4.1, 4.2, and 4.3).
You can copy and paste this directly into an Overleaf project.
\documentclass[12pt, a4paper]article
\usepackage[utf8]inputenc
\usepackagegeometry
\usepackageamsmath, amssymb, amsthm
\usepackageenumitem
\geometrymargin=1in
% Theorem Styles
\newtheorempropositionProposition
\newtheoremproblemProblem
\titleSolutions to Dummit \& Foote: Chapter 4\\Group Actions
\authorCompiled Solutions
\date\today
\begindocument
\maketitle
\sectionSection 4.1: Group Actions and Permutation Representations
\beginproblem[Exercise 4.1.1]
Let $G$ be a group acting on a set $A$. Prove that the relation $\sim$ defined by $a \sim b$ if and only if $b = g \cdot a$ for some $g \in G$ is an equivalence relation.
\endproblem
\beginproof
To show $\sim$ is an equivalence relation, we must verify reflexivity, symmetry, and transitivity.
\beginenumerate[label=(\roman*)]
\item \textbfReflexivity: Let $a \in A$. Since $G$ acts on $A$, $1 \cdot a = a$ for the identity element $1 \in G$. Thus, $a \sim a$.
\item \textbfSymmetry: Suppose $a \sim b$. Then there exists $g \in G$ such that $b = g \cdot a$. Since $G$ is a group, $g^-1 \in G$. Then:
\[ g^-1 \cdot b = g^-1 \cdot (g \cdot a) = (g^-1g) \cdot a = 1 \cdot a = a. \]
Thus, $a = g^-1 \cdot b$, which implies $b \sim a$.
\item \textbfTransitivity: Suppose $a \sim b$ and $b \sim c$. Then there exist $g, h \in G$ such that $b = g \cdot a$ and $c = h \cdot b$. Substituting, we get:
\[ c = h \cdot (g \cdot a) = (hg) \cdot a. \]
Since $hg \in G$, we have $a \sim c$.
\endenumerate
\endproof
\beginproblem[Exercise 4.1.3]
Show that the stabilizer $G_a$ of a point $a$ is a subgroup of $G$.
\endproblem
\beginproof
Let $G_a = \g \in G \mid g \cdot a = a\$.
\beginenumerate[label=(\roman*)]
\item \textbfIdentity: Since $1 \cdot a = a$, $1 \in G_a$.
\item \textbfClosed under inverses: If $g \in G_a$, then $g \cdot a = a$. Applying $g^-1$ to both sides:
\[ g^-1 \cdot (g \cdot a) = g^-1 \cdot a \implies 1 \cdot a = g^-1 \cdot a \implies a = g^-1 \cdot a. \]
Thus, $g^-1 \in G_a$.
\item \textbfClosed under products: If $g, h \in G_a$, then:
\[ (gh) \cdot a = g \cdot (h \cdot a) = g \cdot a = a. \]
Thus, $gh \in G_a$.
\endenumerate
Therefore, $G_a \le G$.
\endproof
\sectionSection 4.2: The Class Equation
\beginproblem[Exercise 4.2.1]
Let $G$ be a finite group of order $n$. Show that the size of the conjugacy class of an element $x \in G$ divides $n$.
\endproblem
\beginproof
The group $G$ acts on itself by conjugation. The orbit of an element $x$ under this action is its conjugacy class, denoted $\mathcalO_x$ or $\textCl(x)$. The stabilizer of $x$ is the centralizer $C_G(x) = \g \in G \mid gxg^-1 = x\$.
By the Orbit-Stabilizer Theorem:
\[ |\mathcalO_x| = [G : C_G(x)]. \]
The index $[G : C_G(x)]$ divides $|G| = n$ by Lagrange's Theorem. Therefore, the size of the conjugacy class divides $n$.
\endproof
\sectionSection 4.3: Group Actions on Sets
\beginproblem[Exercise 4.3.5]
Show that if $G$ is a group of order $p^2$ ($p$ prime), then $G$ is abelian.
\endproblem
\beginproof
The center of $G$, denoted $Z(G)$, is non-trivial for any $p$-group. Thus $|Z(G)|$ is either $p$ or $p^2$.
\beginenumerate
\item Suppose $|Z(G)| = p^2$. Then $Z(G) = G$, so $G$ is abelian.
\item Suppose $|Z(G)| = p$. Then the order of the quotient $G/Z(G)$ is $p$. Groups of prime order are cyclic. Let $G/Z(G) = \langle xZ(G) \rangle$.
Let $g, h \in G$. Then $gZ(G) = x^iZ(G)$ and $hZ(G) = x^jZ(G)$ for some $i,j$. This implies $g = x^i z_1$ and $h = x^j z_2$ for $z_1, z_2 \in Z(G)$.
Since elements in $Z(G)$ commute with everyone:
\[ gh = (x^i z_1)(x^j z_2) = x^i+j z_1 z_2. \]
\[ hg = (x^j z_2)(x^i z_1) = x^j+i z_2 z_1. \]
Since $x^i+j = x^j+i$ and $z_1 z_2 = z_2 z_1$, we have $gh = hg$. Thus $G$ is abelian.
\endenumerate
In either case, $G$ is abelian.
\endproof
\enddocument
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\documentclass[12pt]article
\usepackage[utf8]inputenc
\usepackageamsmath, amssymb, amsthm
\usepackageenumitem
\usepackage[margin=1in]geometry
\titleDummit \& Foote \\ Chapter 4: Group Actions \\ Solutions
\authorOverleaf Write-up
\date{}
\begindocument
\maketitle
\section*Section 4.1: Group Actions and Permutation Representations
\subsection*Exercise 1
Let $G$ act on the set $A$. Prove that for each fixed $g \in G$, the map $\sigma_g : A \to A$ defined by $\sigma_g(a) = g \cdot a$ is a permutation of $A$.
\beginproof
We show $\sigma_g$ is bijective.
\textitInjectivity: If $\sigma_g(a)=\sigma_g(b)$, then $g\cdot a = g\cdot b$. Multiply by $g^-1$ on the left (using the action axioms): $a = e\cdot a = g^-1\cdot(g\cdot a) = g^-1\cdot(g\cdot b) = b$.
\textitSurjectivity: For any $b\in A$, let $a = g^-1\cdot b$. Then $\sigma_g(a)=g\cdot(g^-1\cdot b)=b$.
Thus $\sigma_g \in S_A$.
\endproof
\subsection*Exercise 2
Show that the map $\varphi: G \to S_A$ given by $\varphi(g)=\sigma_g$ is a group homomorphism.
\beginproof
For $g,h \in G$ and $a\in A$:
\[
\varphi(gh)(a) = (gh)\cdot a = g\cdot(h\cdot a) = \sigma_g(\sigma_h(a)) = (\sigma_g \circ \sigma_h)(a) = (\varphi(g)\varphi(h))(a).
\]
Hence $\varphi(gh)=\varphi(g)\varphi(h)$.
\endproof
\subsection*Exercise 3
Let $G$ act on $A$. Prove that the kernel of the homomorphism $\varphi: G\to S_A$ is $\bigcap_a\in A G_a$, where $G_a = \g \in G \mid g\cdot a = a\$ is the stabilizer of $a$.
\beginproof
\[
g \in \ker\varphi \iff \varphi(g)=\textid_A \iff g\cdot a = a \ \forall a\in A \iff g \in \bigcap_a\in A G_a.
\]
\endproof
\subsection*Exercise 4
Let $G$ be a group of order $n$ acting on a set $A$ of size $m$. Show that the kernel of the action is a normal subgroup of $G$ and that $G/\ker\varphi$ is isomorphic to a subgroup of $S_m$.
\beginproof
$\ker\varphi$ is a normal subgroup (kernel of homomorphism). By the First Isomorphism Theorem, $G/\ker\varphi \cong \operatornameIm\varphi \le S_m$.
\endproof
\subsection*Exercise 5
Let $G$ act on $A$ and fix $a\in A$. Prove that $G_a \le G$ and for any $g\in G$, $G_g\cdot a = g G_a g^-1$.
\beginproof
$G_a$ contains identity and is closed under multiplication and inverses. For the second part:
\[
h \in G_g\cdot a \iff h\cdot(g\cdot a) = g\cdot a \iff (g^-1hg)\cdot a = a \iff g^-1hg \in G_a \iff h \in g G_a g^-1.
\]
\endproof
\section*Section 4.2: Orbits and Stabilizers
\subsection*Exercise 6
Let $G$ act on $A$. Define $a\sim b$ if $b = g\cdot a$ for some $g\in G$. Show this is an equivalence relation.
\beginproof
\textitReflexive: $a = e\cdot a$.
\textitSymmetric: $b=g\cdot a \implies a = g^-1\cdot b$.
\textitTransitive: $b=g\cdot a, c=h\cdot b \implies c = (hg)\cdot a$.
\endproof
\subsection*Exercise 7
State and prove the Orbit–Stabilizer Theorem.
\begintheorem[Orbit–Stabilizer]
Let $G$ act on $A$ and $a\in A$. Then $|\mathcalO_a| = [G : G_a]$, where $\mathcalO_a = \g\cdot a \mid g\in G\$.
\endtheorem
\beginproof
Define $\psi: G/G_a \to \mathcalO_a$ by $\psi(gG_a)=g\cdot a$. Well-defined: $gG_a = hG_a \iff h^-1g\in G_a \iff (h^-1g)\cdot a = a \iff g\cdot a = h\cdot a$. $\psi$ is bijective (surjective by definition, injective by the previous equivalence). Hence $|\mathcalO_a| = |G/G_a| = [G:G_a]$.
\endproof
\subsection*Exercise 8
Let $G$ be a finite group acting on a finite set $A$. Prove Burnside's Lemma: The number of orbits is $\frac1\sum_g\in G |\operatornameFix(g)|$, where $\operatornameFix(g)=\a\in A \mid g\cdot a = a\$.
\beginproof
Count pairs $(g,a)$ with $g\cdot a = a$ in two ways:
$\sum_g\in G|\operatornameFix(g)| = \sum_a\in A|G_a|$.
By Orbit–Stabilizer, $|G_a| = |G|/|\mathcalO_a|$. Hence
\[
\sum_a\in A \frac = |G| \sum_\textorbits O \sum_a\in O \frac1O = |G| \cdot (\text\# orbits).
\]
Dividing by $|G|$ gives the result.
\endproof
\subsection*Exercise 9
Let $G$ be a group of order $p^k$ ($p$ prime) acting on a finite set $A$. Show that $|A| \equiv |\operatornameFix(G)| \pmodp$, where $\operatornameFix(G)=\a\in A \mid g\cdot a = a \ \forall g\in G\$.
\beginproof
Write $A$ as a disjoint union of orbits. Each nontrivial orbit has size dividing $|G|$, hence divisible by $p$. Thus $|A| \equiv |\operatornameFix(G)| \pmodp$.
\endproof
\section*Section 4.3: Examples of Group Actions
\subsection*Exercise 10
Let $G$ act on itself by left multiplication. Show that this action is faithful and transitive.
\beginproof
Faithful: If $g\cdot h = h$ for all $h\in G$, then $g=e$.
Transitive: For any $h_1,h_2$, let $g = h_2h_1^-1$ gives $g\cdot h_1 = h_2$.
\endproof
\subsection*Exercise 11
Let $G$ act on itself by conjugation: $g\cdot x = gxg^-1$. Determine the orbits (conjugacy classes) and stabilizer (centralizer $C_G(x)$).
\beginproof
Orbit: $\gxg^-1 \mid g\in G\$. Stabilizer: $\g\in G \mid gxg^-1=x\ = C_G(x)$.
Orbit–Stabilizer gives $| \textconjugacy class of x | = [G : C_G(x)]$.
\endproof
\subsection*Exercise 12
Let $G$ act on the set of subgroups by conjugation: $g\cdot H = gHg^-1$. Show that the stabilizer of $H$ is the normalizer $N_G(H)$.
\beginproof
$g\in \operatornameStab(H) \iff gHg^-1=H \iff g\in N_G(H)$.
\endproof
\section*Section 4.4: The Sylow Theorems (Statement and Applications)
\subsection*Exercise 13
State the three Sylow theorems.
\beginenumerate[label=(\roman*)]
\item For any prime $p$ dividing $|G|$, $G$ has a Sylow $p$-subgroup (of order $p^a$ where $p^a \mid |G|$ but $p^a+1\nmid |G|$).
\item All Sylow $p$-subgroups are conjugate. The number $n_p$ of Sylow $p$-subgroups satisfies $n_p \equiv 1 \pmodp$ and $n_p \mid |G|/p^a$.
\item Any $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.
\endenumerate
\subsection*Exercise 14
Let $|G|=pq$ with primes $p<q$ and $p \nmid q-1$. Show $G$ is cyclic.
\beginproof
By Sylow, $n_q \equiv 1 \pmodq$ and $n_q \mid p$, so $n_q=1$. Thus the Sylow $q$-subgroup $Q$ is normal. $n_p \equiv 1 \pmodp$ and $n_p \mid q$, so $n_p=1$ (since $p<q$ and $p\nmid q-1$ forces $n_p\neq q$). Hence $G$ is direct product of cyclic groups of orders $p$ and $q$, which are coprime, so $G\cong C_pq$ cyclic.
\endproof
\subsection*Exercise 15
Prove that there is no simple group of order $56 = 2^3\cdot 7$.
\beginproof
$n_7 \equiv 1 \pmod7$ and $n_7 \mid 8$, so $n_7=1$ or $8$. If $n_7=1$, the Sylow $7$-subgroup is normal. If $n_7=8$, then $8(7-1)=48$ elements of order $7$. The remaining $56-48=8$ elements form the Sylow $2$-subgroups; each Sylow $2$-subgroup has order $8$. But then $n_2 \mid 7$ and $n_2\equiv 1 \pmod2$, so $n_2=1$ or $7$. $n_2=1$ gives a normal subgroup. $n_2=7$ gives $7$ subgroups of order $8$, each containing identity, total elements $7\cdot 7 +1$? Let's check carefully: the intersection of distinct Sylow $2$-subgroups can be large; but a standard argument: if $n_7=8$, then the normalizer of a Sylow $7$ has index $8$, so $|N_G(P_7)|=7$. But $P_7$ is cyclic of order $7$, so $N_G(P_7)$ contains $P_7$ and possibly an element of order $2$ (since $56/7=8$, the normalizer size is $7$ or $56$; if $n_7=8$, then $|N_G(P_7)|=7$, so no element of order $2$ normalizes $P_7$, contradiction to counting). Thus $n_7$ cannot be $8$. Hence $n_7=1$, so $G$ not simple.
\endproof
\section*Section 4.5: Applications to Finite Groups
\subsection*Exercise 16
Let $G$ be a non‑abelian group of order $p^3$ ($p$ prime). Prove $|Z(G)|=p$.
\beginproof
$Z(G)$ is nontrivial by class equation. $|Z(G)|$ divides $p^3$, so possible $p, p^2, p^3$. If $|Z(G)|=p^3$, $G$ abelian, contradiction. If $|Z(G)|=p^2$, then $G/Z(G)$ is cyclic of order $p$, implying $G$ abelian (since if $G/Z$ cyclic then $G$ abelian), contradiction. Hence $|Z(G)|=p$.
\endproof
\subsection*Exercise 17
Show that a group of order $p^2$ ($p$ prime) is abelian.
\beginproof
$|Z(G)|>1$ by class equation. So $|Z(G)|=p$ or $p^2$. If $p$, then $G/Z(G)$ has order $p$, hence cyclic, so $G$ abelian (contradiction to $|Z(G)|=p$ unless $G$ abelian). Wait careful: If $|Z(G)|=p$, then $G/Z(G)$ cyclic $\implies G$ abelian $\implies Z(G)=G$, so $|Z(G)|=p^2$. So the only possibility is $|Z(G)|=p^2$, i.e., $G$ abelian.
\endproof
\subsection*Exercise 18
Let $G$ act transitively on $A$ with $|A|>1$. Show there exists $g\in G$ with no fixed points (i.e., $\operatornameFix(g)=\emptyset$).
\beginproof
By Burnside's Lemma, number of orbits $=1 = \frac1\sum_g\in G|\operatornameFix(g)|$. So $\sum_g\in G|\operatornameFix(g)| = |G|$. If every $g\neq e$ had at least one fixed point, then $|\operatornameFix(e)|=|A|>1$ gives total sum $>|G|$ (since $|A| + (|G|-1)\cdot 1 > |G|$). Contradiction. Hence some non‑identity element has no fixed points.
\endproof
\section*Section 4.6: Actions on the Coset Space and the Class Equation
\subsection*Exercise 19
Let $H\le G$. Show that the action of $G$ on the left cosets $G/H$ by left multiplication is transitive with kernel $\bigcap_x\in G xHx^-1$.
\beginproof
Transitive: For any $aH, bH$, $(ba^-1)\cdot aH = bH$.
Kernel: $g\in \ker \iff gxH = xH \ \forall x \iff x^-1gx \in H \ \forall x \iff g \in \bigcap_x\in G xHx^-1$.
\endproof
\subsection*Exercise 20
State the class equation for a finite group $G$:
\[
|G| = |Z(G)| + \sum [G : C_G(g_i)],
\]
where the sum runs over representatives of conjugacy classes of size $>1$.
\beginproof
$G$ is the union of its conjugacy classes. The size of the class of $g$ is $[G:C_G(g)]$. The center $Z(G)$ consists of classes of size $1$.
\endproof
\subsection*Exercise 21
Prove that if $|G|=p^n$ for $p$ prime, then $Z(G)\neq 1$.
\beginproof
From class equation, $|G| = |Z(G)| + \sum [G:C_G(g_i)]$. Each $[G:C_G(g_i)]$ is a power $p^k_i$ with $k_i\ge 1$ for non‑central elements. Hence $|Z(G)| = p^n - \sum p^k_i$ is divisible by $p$, so $|Z(G)|\ge p$.
\endproof
\section*Appendix: Selected Additional Exercises
\subsection*Exercise 22 (4.3.7)
Let $G$ act on $A$ and let $a,b\in A$ be in the same orbit. Prove $|G_a|=|G_b|$.
\beginproof
$b = g\cdot a$, so $G_b = gG_ag^-1$, hence isomorphic and same cardinality.
\endproof
\subsection*Exercise 23 (4.4.12)
Show that a group of order $30$ has a normal Sylow $5$-subgroup.
\beginproof
$n_5 \equiv 1 \pmod5$ and $n_5 \mid 6$, so $n_5=1$ or $6$. If $n_5=6$, then there are $6(5-1)=24$ elements of order $5$. Then $n_3 \equiv 1 \pmod3$ and $n_3 \mid 10$, so $n_3=1$ or $10$. $n_3=10$ gives $20$ elements of order $3$, total $24+20=44 >30$, impossible. Hence $n_3=1$ (normal Sylow $3$). The Sylow $5$ and Sylow $3$ intersect trivially, so $G$ has a normal subgroup of order $15$, which contains a unique Sylow $5$, so $n_5=1$.
\endproof
\section*Conclusion
These solutions cover the core ideas of Chapter 4: group actions, orbits, stabilizers, Burnside’s lemma, Sylow theorems, class equation, and their applications to classifying finite groups. Each proof emphasizes the constructive use of actions to reduce group‑theoretic problems to counting arguments.
\enddocument
A comprehensive LaTeX template for Dummit & Foote Chapter 4 solutions on Overleaf requires structuring around Group Actions and Sylow Theorems, utilizing amsmath, amssymb, and amsthm packages for mathematical rigor. Key features for managing complex algebraic proofs include using the proof environment, implementing hyperref for navigation, and using TikZ for diagramming group orbits.
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Chapter 4 of Dummit and Foote’s Abstract Algebra focuses on Group Actions, covering foundational concepts like the Orbit-Stabilizer Theorem, Sylow's Theorems, and the Simplicity of Ancap A sub n
. Complete solutions for this chapter are often sought after for graduate-level qualifying exam prep and course homework. Overview of Chapter 4 Content Chapter 4 exercises typically revolve around:
Section 4.1-4.2: Basic definitions of group actions, orbits, and stabilizers. Exercises often require verifying the action properties or calculating specific stabilizers.
Section 4.3: Groups acting on themselves by conjugation (the Class Equation). Section 4.4: Automorphisms and the action of on its subgroups.
Section 4.5: Sylow's Theorems, which are critical for proving a group is not simple. Finding Solutions on Overleaf
While Overleaf is a LaTeX editor and not a content repository, many students and educators host their Dummit and Foote solution projects there or share the source code on platforms like GitHub to be imported into Overleaf. Greg Kikola's Solutions
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Example pattern: "Find the conjugacy classes of $S_4$ and verify the class equation."
Solution strategy: List cycle types, compute centralizer sizes, then verify $|G| = |Z(G)| + \sum [G : C_G(g_i)]$. Use a table in LaTeX (\begintabular) to present classes cleanly.