hcju/proofwikiqa huggingface dataset:
First note that from Group Homomorphism Preserves Inverses: :$\map \theta {g_2}^{-1} = \paren {\map \theta {g_2} }^{-1} = \map \theta { {g_2}^{-1} }$ and so $\map \theta {g_1} \map \theta {g_2}^{-1}$ is not ambiguous: :$\map \theta {g_1} \map \theta {g_2}^{-1} = \map \theta {g_1} \paren {\map \theta {g_2} }^{-1} = \map \theta {g_1} \map \theta { {g_2}^{-1} }$ From External Direct Product of Groups is Group, $G \times G$ is a group. Let $a_1, a_2, b_1, b_2 \in G$. We have: {{begin-eqn}} {{eqn | l = \map \phi {a_1 b_1, a_2 b_2}       | r = \map \theta {a_1 b_1} \map \theta {a_2 b_2}^{-1}       | c =  }} {{eqn | r = \map \theta {a_1 b_1} \paren {\map \theta {a_2 b_2} }^{-1}       | c = Group Homomorphism Preserves Inverses }} {{eqn | r = \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {a_2} \map \theta {b_2} }^{-1}       | c = {{Defof|Group Homomorphism}} }} {{eqn | r = \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {b_2}^{-1} } \paren {\map \theta {a_2}^{-1} }       | c = Inverse of Group Product }} {{eqn | r = \map \theta {a_1} \paren {\map \theta {a_2}^{-1} } \map \theta {b_1} \paren {\map \theta {b_2}^{-1} }       | c = {{Defof|Abelian Group}}: $H$ is Abelian }} {{eqn | r = \map \phi {a_1, a_2} \map \phi {b_1, b_2}       | c = Definition of $\phi$ }} {{end-eqn}} Hence the result by definition of homomorphism. {{Qed}}
{{AimForCont}} $\map {\rho^k} i = \map {\rho^r} i$ for some $r > 0$. As $\rho$ has an inverse in $S_n$: :$\map {\rho^{k - r} } i = i$ This contradicts the definition of $k$, because $k - r < k$ Thus: :$r = 0$ The result follows. {{qed}}
Let $x \in R$ and $\epsilon \in \R_{\gt 0}$ Then for $y \in R$: {{begin-eqn}} {{eqn | l = \norm {y - x}_1 < \epsilon       | o = \leadstoandfrom       | r = \norm {y - x}_2^\alpha < \epsilon }} {{eqn | o = \leadstoandfrom       | r = \norm {y - x}_2 < \epsilon^{1 / \alpha} }} {{end-eqn}} Hence: :$\map {B^1_\epsilon} x = \map {B^2_{\epsilon^{1 / \alpha} } } x$ where: :$\map {B^1_\epsilon} x$ is the open ball in $d_1$ centered on $x$ with radius $\epsilon$ :$\map {B^2_{\epsilon^{1 / \alpha} } } x$ is the open ball in $d_2$ centered on $x$ with radius $\epsilon^{1 / \alpha}$ Since $x$ and $\epsilon$ were arbitrary then: :every $d_1$-open ball is a $d_2$-open ball. Similarly, for $y \in R$: {{begin-eqn}} {{eqn | l = \norm {y - x}_2 < \epsilon       | o = \leadstoandfrom       | r = \norm {y - x}_2^\alpha < \epsilon^\alpha }} {{eqn | o = \leadstoandfrom       | r = \norm {y - x}_1 < \epsilon^\alpha }} {{end-eqn}} So: :every $d_2$-open ball is a $d_1$-open ball. By the definition of an open set of a metric space it follows that $d_1$ and $d_2$ are topologically equivalent metrics, {{qed}}
The contrapositive is proved. Let there exist $x \in R$ such that $\norm x_1 < 1$ and $\norm x_2 \ge 1$. Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$. By Sequence of Powers of Number less than One in Normed Division Ring then $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1$. By convergent sequence in normed division ring is a Cauchy sequence then $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$. Let $0_R$ be the zero of $R$ and $1_R$ be the unit of $R$. By Norm of Unity of Division Ring and the assumption $\norm x_1 < 1$: :$x \ne 1_R$ Hence: :$x - 1_R \ne 0_R$ By norm axiom $(\text N 1)$: Positive Definiteness: :$\norm {x - 1_R}_2 > 0$ Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$. Then $\norm {x - 1_R}_2 > \epsilon$. We have that $\norm x_2 \ge 1$. Hence for all $n \in \N$: {{begin-eqn}} {{eqn | l = \norm {x_n}_2       | r = \norm {x^n}_2       | c = Definition of $x_n$ }} {{eqn | r = \norm x_2^n       | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = \ge       | r = 1       | c =  }} {{end-eqn}} For all $n \in \N$: {{begin-eqn}} {{eqn | l = \norm {x_{n + 1} - x_n}_2       | r = \norm {x^{n + 1} - x^n}_2 }} {{eqn | r = \norm {x^n x - x^n}_2 }} {{eqn | r = \norm {x^n \paren {x - 1_R} }_2 }} {{eqn | r = \norm {x^n}_2 \norm {x - 1_R}_2       | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = >       | r = \epsilon }} {{end-eqn}}  So $\sequence {x_n}$ is not a Cauchy sequence in $\norm {\, \cdot \,}_2$. The theorem now follows by the Rule of Transposition.  {{qed}}
Consider the dihedral group $D_4$, whose group presentation is: {{:Group Presentation of Dihedral Group D4}} We have that: :$\order {a^2} = 2$ and: :$\order b = 2$ but $a^2$ and $b$ are not conjugate to each other. {{qed}}
Let $\rho, \sigma \in S_n$. Let $v = \displaystyle \sum_{k \mathop = 1}^n \lambda_k v_k$. We have: {{begin-eqn}} {{eqn | l = \rho * \paren {\sigma * v}       | r = \rho * \paren {\sigma * \sum_{k \mathop = 1}^n \lambda_k v_k}       | c = Definition of $v$ }} {{eqn | r = \rho * \sum_{k \mathop = 1}^n \lambda_k v_{\map \sigma k}       | c = Definition of $*$ }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho {\map \sigma k} }       | c = Definition of $*$ }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map {\rho \mathop \circ \sigma} k}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \paren {\rho \circ \sigma} * \sum_{k \mathop = 1}^n \lambda_k v_k       | c = Definition of $*$ }} {{eqn | r = \paren {\rho \circ \sigma} * v       | c = Definition of $v$ }} {{end-eqn}} Hence $*$ fulfils {{GroupActionAxiom|1}}. Let $e$ denote the identity element of $S_n$. Then: {{begin-eqn}} {{eqn | l = e * v       | r = e * \sum_{k \mathop = 1}^n \lambda_k v_k       | c = Definition of $v$ }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map e k}       | c = Definition of $*$ }} {{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_k       | c = Definition of $e$ }} {{eqn | r = v       | c = Definition of $v$ }} {{end-eqn}} Hence $*$ fulfils {{GroupActionAxiom|2}}. {{qed}}
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_1$. Let $\epsilon > 0$ be given. Since $\sequence {x_n}$ is a Cauchy sequence then: :$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$ Then: :$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$ Hence: :$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$ So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$ It follows that for all sequences $\sequence {x_n}$ in $R$: :$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$ {{qed|lemma}} Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_2$. Let $\epsilon > 0$ be given. Since $\sequence {x_n}$ is a Cauchy sequence then: :$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$ Then: :$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$ Hence: :$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$ So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ It follows that for all sequences $\sequence {x_n}$ in $R$: :$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ The result follows. {{qed}}
Let $\sigma \in S_n$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$. Let $\rho$ be conjugate to $\sigma$ From Conjugate Permutations have Same Cycle Type, $\rho$ has the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$ as $\sigma$. That is, all the elements of the same conjugacy class have the same cycle type. {{qed|lemma}} Let $\sigma, \rho \in S_n$ have the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$. It is to be demonstrated that $\sigma$ and $\rho$ are in the same conjugacy class. From Existence and Uniqueness of Cycle Decomposition, $\sigma$ and $\rho$ can each be expressed uniquely as the product of disjoint cycles: {{begin-eqn}} {{eqn | l = \sigma       | r = \alpha_1 \alpha_2 \dotsm \alpha_l }} {{eqn | l = \rho       | r = \beta_1 \beta_2 \dotsm \beta_l }} {{end-eqn}} where $\alpha_i$ and $\beta_i$ are $k_i$-cycles. For each $i$, let the $k_i$-cycles $\alpha_i$ and $\beta_i$ be expressed as: {{begin-eqn}} {{eqn | l = \alpha_i       | r = \alpha_{i1} \alpha_{i2} \dotsm \alpha_{i k_l} }} {{eqn | l = \beta_i       | r = \beta_{i1} \beta_{i2} \dotsm \beta_{i k_l} }} {{end-eqn}} For all $i, j$ such that $1 \le i \le l$, $1 \le j \le k_i$, let: :$\tau := \map \tau {\alpha_{i j} } = \beta_{i j}$ Such a $\tau$ is bound to exist in $S_n$, as the underlying set of $S_n$ is the set of all permutations of $\set {1, 2, \ldots, n}$. Thus from Product of Conjugates equals Conjugate of Products: :$\tau \alpha_i \tau^{-1} = \beta_i$ Hence: {{begin-eqn}} {{eqn | l = \tau \sigma \tau^{-1}       | r = \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1} }} {{eqn | r = \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1} }} {{eqn | r = \beta_1 \beta_2 \dotsm \beta_l }} {{eqn | r = \rho }} {{end-eqn}} demonstrating that $\sigma$ and $\rho$ are conjugate. That is, $\sigma$ and $\rho$ are in the same conjugacy class. Hence the result. {{qed}} Category:Symmetric Groups Category:Conjugacy Classes lqejkraisnceer65vizzoa9d6n6znq1
From Center is Intersection of Centralizers: :$\displaystyle \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$ where $\map {C_G} g$ denotes the centralizer of $g$. From Centralizer of Group Element is Subgroup, each of $\map {C_G} g$ is a subgroup of $G$. From Identity of Subgroup: :$\forall g \in G: e \in \map {C_G} g$ Hence by definition of set intersection: :$e \in \displaystyle \bigcap_{g \mathop \in G} \map {C_G} g$ whence the result. {{qed}} Category:Centers of Groups Category:Identity Elements kgz29nmgfr9rmlqsa8b3we40xlppoae
From Identity of Group is in Center: :$e \in \map Z G$ where $\map Z G$ is the center of $G$. From Conjugacy Class of Element of Center is Singleton: :$\conjclass e = \set e$ {{qed}}
Let $G$ be of order $n$. Let $G$ have exactly $2$ conjugacy classes. Let $x \in G$ such that $x \ne e$. Let $\conjclass x$ denote the conjugacy class of $x$. From Identity of Group is in Singleton Conjugacy Class: $\conjclass e = \set e$ where $\conjclass e$ denotes the conjugacy class of $e$ The other elements of $G$ are in $\conjclass x$. Thus: :$\card {\conjclass x} = n - 1$ where $\card {\, \cdot \,}$ denotes the cardinality of a set. We have that Size of Conjugacy Class is Index of Normalizer. From Lagrange's Theorem, the index of any subgroup of $G$ is a divisor of $G$. Thus: :$\paren {n - 1} \divides n$ where $\divides$ denotes divisibility. It follows that $n = 2$. Hence the result by definition of order of group. {{qed}}
Let $G$ be a group of order $15$. We have that $15 = 3 \times 5$. Thus from the First Sylow Theorem: :$G$ has at least one subgroup $H_3$ of order $3$ and: :$G$ has at least one subgroup $H_5$ of order $5$. From Prime Group is Cyclic, all such subgroups of order $3$ and order $5$ are cyclic. {{Qed}}
Let $G$ be a group of order $15$. From the Fourth Sylow Theorem: :the number of Sylow $p$-subgroups is equivalent to $1 \pmod p$ We have that $15 = 3 \times 5$. Hence the result. {{Qed}}
By definition of Sylow $p$-subgroup: :$\order {H_1} = p^r$, where $p^r$ is the highest power of $p$ which is a divisor of $\order {G_1}$. :$\order {H_2} = p^s$, where $p^s$ is the highest power of $p$ which is a divisor of $\order {G_2}$. We have that: :$\order {H_1 \times H_2} = p^{r + s}$ We also have that $p^{r + s}$ is the highest power of $p$ which is a divisor of $\order {G_1 \times G_2}$. Hence it follows by definition that $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$. {{qed}}
From Direct Product of Sylow p-Subgroups is Sylow p-Subgroup, $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$. By Sylow $p$-Subgroup is Unique iff Normal, each of $H_1$ and $H_2$ are normal in $G_1$ and $G_2$ respectively. By Direct Product of Normal Subgroups is Normal, $H_1 \times H_2$ is normal in $G_1 \times G_2$. Again by Sylow $p$-Subgroup is Unique iff Normal, $H_1 \times H_2$ is the unique Sylow $p$-subgroup of $G_1 \times G_2$. {{qed}}
We note that from Intersection of Subgroups is Subgroup that $P \cap H$ is a subgroup of $G$ and also of $H$. Let $G$ be the dihedral group $D_3$, given by its group presentation: {{:Group Presentation of Dihedral Group D3}} By definition of Sylow $p$-subgroup, $\gen a$ is a Sylow $3$-subgroup of $G$. However, $\gen b$ is also a subgroup of $G$, of order $2$. But: :$\gen b \cap \gen a = e$ and $e$ is not a Sylow $3$-subgroup of $G$. {{qed}}
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_p \equiv 1 \pmod p$ From the Fifth Sylow Theorem: :$n_p \divides 2 p$ that is: :$n_p \in \set {1, 2, p, 2 p}$ But $p$ and $2 p$ are congruent to $0$ modulo $p$ So: :$n_p \notin \set {p, 2 p}$ Also we have that $p > 2$. Hence: :$2 \not \equiv 1 \pmod p$ and so it must be that $n_p = 1$. It follows from Sylow $p$-Subgroup is Unique iff Normal that this Sylow $p$-subgroup is normal. {{qed}}
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. From the First Sylow Theorem, there exists at least $1$ Sylow $p$-subgroup of $G$. Let $P$ be such a Sylow $p$-subgroup of $G$. The index of $P$ is $2$. From Subgroup of Index 2 is Normal, $P$ is normal in $G$. From Sylow $p$-Subgroup is Unique iff Normal it follows that there is only $1$ Sylow $p$-subgroup of $G$. {{qed}}
This is a specific instance of Group of Order $p q$ has Normal Sylow $p$-Subgroup, where $q = 2$. {{qed}}
When $p = 2$, the result follows from Groups of Order 4. Let $p$ be an odd prime. From Sylow p-Subgroups of Group of Order 2p, $G$ has exactly $1$ normal subgroup $P$ of order $p$. $p$ is prime number. So from Prime Group is Cyclic, $P$ is a cyclic group. Let $P = \gen x$ for some $x \in G$. By the First Sylow Theorem there exists at least one subgroup of $G$ of order $2$. Hence: :$\exists y \in G: y^2 = e$ It follows that the elements of $G$ are known: :$G = \set {e, x, \ldots, x^{p - 1}, y, y x, y x^{p - 1} }$ Then: {{begin-eqn}} {{eqn | l = y x y^{-1}       | o = \in       | r = P       | c = as $P$ is normal in $G$. }} {{eqn | ll= \leadsto       | l = y x y^{-1}       | r = x^i       | c = for some $i \in \Z_{>0}$ }} {{eqn | ll= \leadsto       | l = y x y^{-1} x       | r = x^{i + 1}       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = y x y x = \paren {y x}^2       | r = x^{i + 1}       | c = $y^2 = e$, so $y = y^{-1}$ }} {{end-eqn}} Thus: :the even powers of $y x$ are powers of $x$ :the odd powers of $y x$ are of the form $y x^j$ for some $j \in \Z_{>0}$. By Order of Element Divides Order of Finite Group: :$\order {y x} \divides 2 p$ where: :$\order {y x}$ denotes the order of $y x$ :$\divides$ denotes divisibility. We have that: :$y x \ne e$ and so by Identity is Only Group Element of Order 1: :$\order {y x} \ne 1$ Thus: :$\order {y x} \in \set {2, p, 2 p}$ {{qed|lemma}} Suppose $i \ne -1$ in $(1)$ above. Then: :$\paren {y x}^2 \ne e$ and so: :$\order {y x} \ne 2$ Because odd powers of $y x$ are of the form $y x^j$: :$\paren {y x}^p \ne e$ and so: :$\order {y x} \ne p$ It follows that: :$\order {y x} = 2 p$ and from Group whose Order equals Order of Element is Cyclic, $G$ is cyclic. Thus, when $i \ne -1$: :$G = \gen {y x}$ and so is cyclic. Thus by Cyclic Group is Abelian: :$y x = x y$ {{qed|lemma}} When $i = -1$ in $(1)$ above, we have that: {{begin-eqn}} {{eqn | l = y x y^{-1} x       | r = e       | c =  }} {{eqn | ll= \leadsto       | l = y x y^{-1}       | r = x^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = y x       | r = x^{-1} y       | c =  }} {{end-eqn}} leading to the group presentation of $G$: :$G = \gen {x, y: x^p = e = y^2, y x = x^{-1} y}$ which is the Group Presentation of Dihedral Group $D_p$. {{qed}}
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$. From the Fourth Sylow Theorem: :$n_p \equiv 1 \pmod p$ From the Fifth Sylow Theorem: :$n_p \divides p q$ where $\divides$ denotes divisibility. The divisors of $p q$ are $1$, $p$, $q$ and $p q$. Of these: :$p$ and $p q$ are $\equiv 0 \pmod p$ and as $p > q$: :$q \equiv q \pmod p$ So the only possibility for $n_p$ is $1$. Let $P$ denote this unique Sylow $p$-subgroups. From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal. {{qed}}
=== Necessary Condition === {{:Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition}}{{qed}}
We calculate the powers of $X$ and $Y$ in turn: {{begin-eqn}} {{eqn | l = X^2       | r = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = X^3       | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}       | c =  }} {{end-eqn}} and so on to: {{begin-eqn}} {{eqn | l = X^7       | r = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}       | c = as $7 \equiv 0 \pmod 7$ }} {{end-eqn}} Thus we have: :$X^7 = \mathbf I$ where $\mathbf I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix. Then: {{begin-eqn}} {{eqn | l = Y^2       | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = Y^3       | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}       | c =  }} {{end-eqn}} Thus we have: :$Y^3 = \mathbf I$ and: :$Y^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$. Then: {{begin-eqn}} {{eqn | l = Y X Y^{-1}       | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 4 & 4 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}       | c =  }} {{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}       | c =  }} {{eqn | r = X^2       | c =  }} {{end-eqn}} and the result is apparent. {{qed}}
Note that a Sylow $2$-subgroup of $G$ is of order $4$. From Sylow $3$-Subgroups in Group of Order 12, there are either $1$ or $4$ Sylow $3$-subgroups. Suppose there is exactly $1$ Sylow $3$-subgroup $P$. Then from Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal. {{qed|lemma}} Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$. Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the trivial subgroup of $G$: :$P_i \cap P_j = \set e$ Thus $G$ contains: :The identity element $e$ :$8$ elements of order $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$ :$3$ more elements, which (along with $e$) must form the Sylow $2$-subgroup of order $4$. This Sylow $2$-subgroup $Q$ must be unique. Hence by Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal. {{qed|lemma}} Hence the result. {{qed}}
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$. From the Fourth Sylow Theorem: :$n_p \equiv 1 \pmod p$ From the Fifth Sylow Theorem: :$n_p \divides p^2 q$ where $\divides$ denotes divisibility. Thus $n_p \in \set {1, q}$. Suppose $p > q$. Then: :$q \not \equiv 1 \pmod p$ and so $n_p \ne q$. Hence $n_p = 1$. From Sylow p-Subgroup is Unique iff Normal it follows that this Sylow $p$-subgroup is normal. {{qed|lemma}} Suppose $q > p$. Then $n_p$ could equal $q$ if $q \equiv 1 \pmod p$. Let $n_q$ denote the number of Sylow $q$-subgroups in $G$. From the Fourth Sylow Theorem: :$n_q \equiv 1 \pmod q$ and so $n_q$ is not a multiple of $q$. From the Fifth Sylow Theorem: :$n_q \divides p^2 q$ It follows that: :$n_q \in \set {1, p, p^2}$ If $n_q = 1$ then it follows from Sylow p-Subgroup is Unique iff Normal that this Sylow $q$-subgroup is normal. {{qed|lemma}} As $p \not \equiv 1 \pmod q$, it follows that $n_q \ne p$. The final possibility to be explored is when $n_q = p^2$. Then: {{begin-eqn}} {{eqn | l = p^2       | o = \equiv       | r = 1       | rr= \pmod q       | c =  }} {{eqn | ll= \leadsto       | l = q       | o = \divides       | r = p^2 - 1       | c =  }} {{eqn | ll= \leadsto       | l = q       | o = \divides       | r = \paren {p + 1} \paren {p - 1}       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = q       | r = p + 1       | c = as $q > p$ }} {{end-eqn}} But then $p$ and $q$ are consecutive prime numbers, and so: :$p = 2, q = 3$ This gives the specific case: :$\order G = 2^2 \times 3$ and the result follows from Normal Sylow p-Subgroups in Group of Order 12. {{qed}}
By Group of Order 15 is Cyclic Group, any subgroup of $G$ of order $15$ is cyclic. It remains to be proved that a subgroup of $G$ of order $15$ exists, and that it is normal. Let $n_3$ denote the number of Sylow $3$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_3 \equiv 1 \pmod 3$ and from the Fifth Sylow Theorem: :$n_3 \divides 10$ where $\divides$ denotes divisibility. It follows that $n_3 \in \set {1, 10}$. Let $n_5$ denote the number of Sylow $5$-subgroups of $G$. From the Fourth Sylow Theorem: :$n_5 \equiv 1 \pmod 5$ and from the Fifth Sylow Theorem: :$n_5 \divides 30$ It follows that $n_5 \in \set {1, 6}$. Suppose $n_3 = 1$. Let $P$ denote the unique Sylow $3$-subgroup of $G$. From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal. Thus we may form the quotient group $G / P$, which is of order $10$. From Groups of Order 2p, $G / P$ has a unique Sylow $5$-subgroup, which we will denote $N / P$. By the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements. As noted, $N$ is cyclic. {{qed|lemma}} Suppose $n_3 = 10$. Let the Sylow $3$-subgroups of $G$ be denoted $P_1, P_2, \ldots, P_{10}$. Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, \ldots, 10}, i \ne j$ is the trivial subgroup of $G$: :$P_i \cap P_j = \set e$ Thus $G$ contains: :The identity element $e$ :$20$ elements of order $3$, of which $2$ each are in $P_1, P_2, \ldots, P_{10}$ :$9$ more elements of $G$ which are in subgroups of $G$ different from the Sylow $3$-subgroups, and so have orders different from $3$. Suppose $n_5 = 6$. By a similar argument to the above, these $6$ Sylow $5$-subgroups contribute $24$ elements of order $5$ to $G$. This is not possible because there are only $9$ elements of $G$ so far unaccounted for. So if $G$ has $10$ Sylow $3$-subgroups, it can have only $1$ Sylow $5$-subgroup, which we will denote $Q$. From Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal. Thus we may form the quotient group $G / Q$, which is of order $16$. From Groups of Order 2p, $G / Q$ has a unique Sylow $3$-subgroup, which we will denote $N / Q$. Again by the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements. As noted, $N$ is cyclic. {{qed|lemma}} Both cases have been accounted for: :that $G$ has $1$ Sylow $3$-subgroup :that $G$ has $10$ Sylow $3$-subgroups. Both cases result in $G$ having a normal subgroup of order $15$ which is cyclic. {{qed}}