\sectionSolutions to Chapter 4 \subsectionProblem 4.1 Problem statement here.\\ Solution: \[ \textYour solution here. \] dummit+and+foote+solutions+chapter+4+overleaf+full

\begintikzcd G \times X \arrow[r, "\textaction"] & X \\ (g, x) \arrow[mapsto, rr] && g\cdot x \endtikzcd Abstract Algebra Finding a complete and well-formatted set

\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 \sectionSolutions to Chapter 4 \subsectionProblem 4

\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