\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 Venture Hub Ninja Legends Mobile Script [UPDATED]
\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\$. Download Indiancouple17zip 27862 Mb Free - 3.79.94.248
\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
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\beginproblem[Exercise 4.3.5] Show that if $G$ is a group of order $p^2$ ($p$ prime), then $G$ is abelian. \endproblem
\sectionSection 4.2: The Class Equation
\beginproblem[Exercise 4.1.3] Show that the stabilizer $G_a$ of a point $a$ is a subgroup of $G$. \endproblem