hcju/proofwikiqa huggingface dataset:
By Group of Order 30 has Normal Cyclic Subgroup of Order 15, $G$ has a normal subgroup of order $15$ which is cyclic. Let this normal cyclic order $15$ subgroup be denoted $N$: :$N = \gen x$ Let $y$ be the generator for any Sylow $2$-subgroup of $G$. Then: {{begin-eqn}} {{eqn | l = y x y^{-1}       | o = \in       | r = N       | c = as $N$ is normal }} {{eqn | ll= \leadsto       | l = y x y^{-1}       | r = x^i       | c = for some $i \in \Z_{\ge 0}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x       | r = y^2 x y^{-2}       | c = as $y^2 = e$ }} {{eqn | r = y \paren {y x y^{-1} } y^{-1}       | c = {{GroupAxiom|1}} }} {{eqn | r = y x^i y^{-1}       | c = as $y x y^{-1} = x^i$ }} {{eqn | r = \paren {y x y^{-1} }^i       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = \paren {x^i}^i       | c = as $y x y^{-1} = x^i$ }} {{eqn | r = x^{i^2}       | c = Powers of Group Elements }} {{end-eqn}} and so: :$i^2 - 1 \equiv 0 \pmod {15}$ Investigating the powers of $i$, case by case, searching for those which satisfy this congruence, yields: :$i \in \set {1, 4, 11, 14}$ The case $i \equiv 1 \pmod {15}$ leads to the cyclic group $C_{30}$. The case where $i \equiv {14} \equiv {-1} \pmod {15}$ leads to the dihedral group $D_{15}$. The other two cases lead to: :$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$ :$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$ {{qed}}
We note that: :$24 = 3 \times 2^3$ Hence a Sylow $2$-subgroup of $G$ is of order $8$. Let $n_2$ denote the number of Sylow $2$-subgroups of $G$. By the Fourth Sylow Theorem: :$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd and from the Fifth Sylow Theorem: :$n_2 \divides 24$ where $\divides$ denotes divisibility. It follows that $n_3 \in \set {1, 3}$. Suppose $n_2 = 1$. Let $P$ denote the unique Sylow $2$-subgroup of $G$. From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal. This is therefore the required normal subgroup of order $8$. {{qed|lemma}} Suppose $n_2 = 3$. Let $S_1$, $S_2$ and $S_3$ denote the $3$ Sylow $2$-subgroups of $G$. Each has order $8$, as noted earlier. Consider the product $S_1 S_2$. By Order of Subgroup Product: :$\order {S_1 S_2} = \dfrac {2^3 2^3} {2^r}$ where: :$\order {S_1 S_2}$ denotes the order of $S_1 S_2$ :$\order {S_1 \cap S_2} = 2^r$ As $S_1 S_2 \subseteq G$ and $\order G = 24$: :$\order {S_1 S_2} \le 24$ Hence: {{begin-eqn}} {{eqn | l = 2^3 2^3       | o = \le       | r = \order {S_1 S_2} \times 2^r       | c =  }} {{eqn | ll= \leadsto       | l = 64       | o = \le       | r = 24 \times 2^r       | c =  }} {{eqn | ll= \leadsto       | l = r       | o = \ge       | r = 2       | c =  }} {{end-eqn}} As $S_1 \cap S_2$ is a proper subgroup of $S_1$, it can have no more than $2^2 = 4$ elements. So if $G$ has $3$ Sylow $2$-subgroups, the intersection of any $2$ of them is of order $4$. Let $T = S_1 \cap S_2$, so that $\order T = 4$. We have that: :$\index T {S_1} = 2$ where $\index T {S_1}$ denotes the index of $T$ in $S_1$. From Subgroup of Index 2 is Normal, $T$ is normal in $S_1$. Similarly, $T$ is normal in $S_2$. Thus $S_1$ and $S_2$ are both subgroups of the normalizer $\map {N_G} T$ of $T$ in $G$. Thus $H = \gen {S_1, S_2}$ is a subgroup of $\map {N_G} T$. Hence $T$ is a normal subgroup of $H$. As $H$ is a subgroup of $\map {N_G} T$, it contains $S_1 S_2$ as a subset. But $S_1 S_2$ contains $\dfrac {2^6} {2^2} = 16$ elements. The only subgroup of $G$ containing $16$ elements has to be $G$ itself. So $H = G$ and so $\map {N_G} T = G$. That is, $T$ is a normal subgroup of $G$. This is therefore the required normal subgroup of order $4$. {{qed}}
We have that $35 = 5 \times 7$. Then we have that $5$ and $7$ are primes such that $5 < 7$ and $5$ does not divide $7 - 1$. Thus Cyclic Groups of Order $p q$ can be applied. {{Qed}}
Let $G$ be of order $35$. We have that $35 = 5 \times 7$ where both $5$ and $7$ are prime. Hence from the First Sylow Theorem, $G$ has: :at least one Sylow $5$-subgroup and: :at least one Sylow $7$-subgroup Let $n_5$ denote the number of Sylow $5$-subgroups of $G$. Let $n_7$ denote the number of Sylow $7$-subgroups of $G$. By the Fourth Sylow Theorem: :$n_5 \equiv 1 \pmod 5$ and from the Fifth Sylow Theorem: :$n_5 \divides 35$ where $\divides$ denotes divisibility. It follows that $n_5 = 1$. Let $P$ denote this unique Sylow $5$-subgroup. By the Fourth Sylow Theorem: :$n_7 \equiv 1 \pmod 7$ and from the Fifth Sylow Theorem: :$n_7 \divides 35$ where $\divides$ denotes divisibility. It follows that $n_7 = 1$. Let $Q$ denote this unique Sylow $7$-subgroup. From Sylow $p$-Subgroup is Unique iff Normal, $P$ and $Q$ are normal subgroups of $G$. Consider $P \cap Q$. By Intersection of Subgroups is Subgroup: :$P \cap Q$ is a subgroup of $P$ :$P \cap Q$ is a subgroup of $Q$ and: :$P \cap Q$ is a subgroup of $G$ By Lagrange's Theorem: :$\order {P \cap Q} \divides \order P = 5$ :$\order {P \cap Q} \divides \order Q = 7$ where: :$\order {P \cap Q}$ denotes the order of $P \cap Q$ :$\divides$ denotes divisibility. Thus: :$\order {P \cap Q} = 1$ and so: :$P \cap Q = \set e$ where $e$ is the identity of $G$. From Prime Group is Cyclic, both $P$ and $Q$ are cyclic. Let: :$P = \gen a$ :$Q = \gen b$ where $\gen a$ denotes the subgroup of $G$ generated by $a$. As $P$ is a normal subgroup of $G$: :$b^{-1} a^{-1} b \in P$ and so as $a \in P$, from {{GroupAxiom|0}}: :$b^{-1} a^{-1} b a \in P$ As $Q$ is a normal subgroup of $G$: :$a^{-1} b a \in Q$ and so $b^{-1} \in Q$, from {{GroupAxiom|0}}: :$b^{-1} a^{-1} b a \in Q$ Thus by definition of intersection: :$b^{-1} a^{-1} b a \in P \cap Q$ But $P \cap Q = \set e$. So: :$b^{-1} a^{-1} b a = e$ and so from Product of Commuting Elements with Inverses :$a b = b a$ From Power of Product of Commutative Elements in Group: :$\paren {a b}^7 = a^7 b^7 = a^7 \ne e$ and: :$\paren {a b}^5 = a^5 b^5 = b^5 \ne e$ By Order of Element Divides Order of Finite Group: :$\order {a b} \divides 35$ where $\order {a b}$ denotes the order of $a b$. Thus $\order {a b} \in \set {1, 5, 7, 35}$ As it has been established that $\order {a b}$ is not $1$, $5$ or $7$, it follows that: :$\order {a b} = 35$ Hence from Group whose Order equals Order of Element is Cyclic, $G$ is a cyclic group. {{Qed}}
Let $G$ be a group of order $105$ whose identity is $e$. We have that: :$105 = 3 \times 5 \times 7$ From the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup, Sylow $5$-subgroup and Sylow $7$-subgroup. Let: :$n_5$ denote the number of Sylow $5$-subgroups of $G$ :$n_7$ denote the number of Sylow $7$-subgroups of $G$. $5$ and $7$ appear in $105$ with multiplicity $1$. Hence any such Sylow $p$-subgroups are prime groups. From Sylow p-Subgroup is Unique iff Normal: :if $n_5 = 1$ then the unique Sylow $5$-subgroup is normal and: :if $n_7 = 1$ then the unique Sylow $7$-subgroup is normal. It remains to be shown that either $n_5 = 1$ or $n_7 = 1$. By the Fourth Sylow Theorem: :$n_5 \equiv 1 \pmod 5$ and from the Fifth Sylow Theorem: :$n_5 \divides 105$ where $\divides$ denotes divisibility. It follows that $n_5 \in \set {1, 21}$. By the Fourth Sylow Theorem: :$n_7 \equiv 1 \pmod 7$ and from the Fifth Sylow Theorem: :$n_7 \divides 105$ where $\divides$ denotes divisibility. It follows that $n_7 \in {1, 15}$. If either $n_5 = 1$ or $n_7 = 1$ the proof is finished. Suppose $n_7 = 15$. As all of these Sylow $7$-subgroups are prime, the intersection of any $2$ of them is $\set e$. Thus, these $15$ Sylow $7$-subgroups contribute $6 \times 15 = 90$ distinct elements to $G$. This leaves $15$ elements still to be accounted for. Suppose $n_5 = 21$. By a similar argument, these $21$ Sylow $5$-subgroups contribute $4 \times 21 = 84$ distinct elements to $G$. None of these elements, apart from $e$, can also be elements of one of the Sylow $7$-subgroups. But there are only $15$ elements available. So it is not possible for both $n_7 = 15$ and $n_5 = 21$. So either $n_7 = 1$ or $n_5 = 1$. The result follows. {{qed}}
Let $G$ be a group of order $105$ whose identity is $e$. From Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup, $G$ has either: :exactly one normal Sylow $5$-subgroup or: :exactly one normal Sylow $7$-subgroup. Suppose $G$ has exactly one normal Sylow $5$-subgroup, which we will denote $P$. Then $G / P$ is the quotient group of $G$ by $P$ and is of order $21$. From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / P$ has exactly one normal Sylow $7$-subgroup which we denote $N / P$. From the Correspondence Theorem, $N$ is a normal subgroup of $G$ of order $7 \times 5 = 35$. It follows from Group of Order 35 is Cyclic Group that $N$ is cyclic. {{qed|lemma}} Now suppose that $G$ has exactly one normal Sylow $7$-subgroup, which we will denote $Q$. Then $G / Q$ is the quotient group of $G$ by $Q$ and is of order $15$. From Group of Order $p q$ has Normal Sylow $p$-Subgroup, $G / Q$ has exactly one normal Sylow $5$-subgroup which we denote $M / Q$. From the Correspondence Theorem, $m$ is a normal subgroup of $G$ of order $5 \times 7 = 35$. It follows from Group of Order 35 is Cyclic Group that $M$ is cyclic. {{qed|lemma}} Both cases have been covered, and the existence of a normal cyclic subgroup of order $35$ has been proved. The result follows by definition of the index of a subgroup. {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \tuple {x, x}       | o = \in       | r = \Delta_S       | c = {{Defof|Diagonal Relation}} }} {{end-eqn}} So $\Delta_S$ is reflexive.
{{begin-eqn}} {{eqn | lo= \forall x, y \in S:       | l = \tuple {x, y}       | o = \in       | r = \Delta_S       | c =  }} {{eqn | lll=\leadsto       | l = x       | r = y       | c = {{Defof|Diagonal Relation}} }} {{eqn | lll=\leadsto       | l = y       | r = x       | c = Equality is Symmetric }} {{eqn | lll=\leadsto       | l = \tuple {y, x}       | o = \in       | r = \Delta_S       | c = {{Defof|Diagonal Relation}} }} {{end-eqn}} So $\Delta_S$ is symmetric.
{{begin-eqn}} {{eqn | lo= \forall x, y, z \in S:       | l = \tuple {x, y}       | o = \in       | r = \Delta_S \land \tuple {y, z} \in \Delta_S       | c =  }} {{eqn | lll=\leadsto       | l = x       | r = y \land y = z       | c = {{Defof|Diagonal Relation}} }} {{eqn | lll=\leadsto       | l = x       | r = z       | c = Equality is Transitive }} {{eqn | lll=\leadsto       | l = \tuple {x, z}       | o = \in       | r = \Delta_S       | c = {{Defof|Diagonal Relation}} }} {{end-eqn}} So $\Delta_S$ is transitive.
Let $G$ be a group of order $56$ whose identity is $e$. We have that: :$56 = 2^3 \times 7$ From the First Sylow Theorem, $G$ has at least one Sylow $2$-subgroup and Sylow $7$-subgroup. Let: :$n_2$ denote the number of Sylow $2$-subgroups of $G$ :$n_7$ denote the number of Sylow $7$-subgroups of $G$. From Sylow p-Subgroup is Unique iff Normal: :if $n_2 = 1$ then the unique Sylow $2$-subgroup is normal and: :if $n_7 = 1$ then the unique Sylow $7$-subgroup is normal. It remains to be shown that either $n_2 = 1$ or $n_7 = 1$. By the Fourth Sylow Theorem: :$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd) and from the Fifth Sylow Theorem: :$n_2 \divides 56$ where $\divides$ denotes divisibility. It follows that $n_2 \in \set {1, 7}$. By the Fourth Sylow Theorem: :$n_7 \equiv 1 \pmod 7$ and from the Fifth Sylow Theorem: :$n_7 \divides 56$ where $\divides$ denotes divisibility. It follows that $n_7 \in {1, 8}$. If either $n_2 = 1$ or $n_7 = 1$ the proof is finished. Suppose $n_7 = 8$. As all of these Sylow $7$-subgroups are prime, the intersection of any $2$ of them is $\set e$. Thus, these $8$ Sylow $7$-subgroups contribute $6 \times 8 = 48$ distinct elements to $G$. This leaves $7$ elements still to be accounted for. A Sylow $2$-subgroup is of order $2^3 = 8$. Thus all $7$ of these remaining elements must all be in that one Sylow $2$-subgroup So if $n_7 \ne 1$, then $n_2 = 1$. The result follows. {{qed}}
Let $G = H = C_2$, the cyclic group of order $2$. Let $G = \gen x$ and $H = \gen y$, so that: :$G = \set {e_G, x}$ :$H = \set {e_H, y}$ where $e_G$ and $e_H$ are the identity elements of $G$ and $H$ respectively. Consider the element $\tuple {x, y} \in G \times H$. We have that: :$\gen {\tuple {x, y} } =\set {\tuple {e_G, e_H}, \tuple {x, y} }$ but this is not the direct product of a subgroup of $G$ with a subgroup of $H$. {{qed}}
It is to be shown that: {{:Definition:Internal Group Direct Product/General Definition/Definition 2}} by definition 2 of Internal Group Direct Product. Condition $(3)$ already gives that $H_i$ is normal for all $i \in \set {1, 2, \ldots, n}$. Condition $(1)$ gives us that each element $g$ of $G$ can be expressed in the form: :$g = h_1 h_2 \dotsm h_n$ with $h_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$. It is now shown that this expression is unique. Suppose that: :$g = h_1 h_2 \dotsm h_n = k_1 k_2 \dotsm k_n$ where $h_i, k_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$ and $h_j \ne k_j$ for at least one $j$. Let $j$ be the largest integer such that $h_j \ne k_j$, so that $h_i = k_i$ for $i > j$. Cancelling $h_i$ for $i > j$ gives: :$h_i h_2 \dotsm h_j = k_1 k_2 \dotsm k_j$ and so: :$h_j {k_j}^{-1} = \paren {h_1 h_2 \dotsm h_{j - 1} }^{-1} \paren {k_1 k_2 \dotsm k_{j - 1} } \in \paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j$ But by condition $(2)$: :$\paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j = \set e$ by definition of independent subgroups. Thus $h_j = k_j$, which contradicts our assertion that $h_j \ne k_j$. Hence the decomposition is unique. {{qed}}
First we introduce a lemma: === Lemma === {{:Groups of Order 30/Lemma}}{{qed|lemma}} From the lemma, it remains to be shown that the group presentations: :$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$ :$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$ give the groups $C_5 \times D_3$ and $C_3 \times D_5$. === Direct Product $C_5 \times D_3$ === {{:Groups of Order 30/C 5 x D 3}} === Direct Product $C_3 \times D_5$ === {{:Groups of Order 30/C 3 x D 5}}
Let $\norm {\,\cdot\,}$ be non-Archimedean. Then by the definition of a non-Archimedean norm, for $n \in \N$: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = \norm {n \cdot 1_R}       | r = \norm {1_R + \dots + 1_R}       | c = ($n$ summands) }} {{eqn | o = \le       | r = \max \set {\norm {1_R}, \ldots, \norm {1_R} }       | c =  }} {{eqn | r = 1       | c = because $\norm {1_R} = 1$ }} {{end-eqn}}
Let: :$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$ Let $x, y \in R$. Let $y = 0_R$ where $0_R$ is the zero of $R$. Then $\norm {x + y} = \norm x = \max \set {\norm x, 0} = \max \set {\norm x, \norm y}$ ==== Lemma 1 ==== {{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 1}} {{qed|lemma}} Hence to complete the proof it is sufficient to prove: :$\forall x \in R: \norm {x + 1_R} \le \max \set {\norm x, 1}$ For $n \in \N$: {{begin-eqn}} {{eqn | l = \norm {x + 1_R}^n       | r = \norm {\sum_{i \mathop = 0}^n \binom n i \cdot x^i}       | c = Binomial Theorem }} {{eqn | o = \le       | r = \sum_{i \mathop = 0}^n \norm {\binom n i \cdot x^i}       | c = {{NormAxiom|3}} }} {{eqn | r = \sum_{i \mathop = 0}^n \norm {\binom n i \cdot 1_R} \norm x^i       | c = {{NormAxiom|2}} }} {{eqn | o = \le       | r = \sum_{i \mathop = 0}^n \norm x^i       | c = $\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$ }} {{end-eqn}} ==== Lemma 2 ==== {{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2}} {{qed|lemma}} Hence {{begin-eqn}} {{eqn | l = \norm {x + 1_R}^n       | o = \le       | r = \sum_{i \mathop = 0}^n \norm x^i       | c = continuing from above }} {{eqn | o = \le       | r = \sum_{i \mathop = 0}^n \max \set {\norm x^n , 1}       | c = Lemma 2 }} {{eqn | r = \paren {n + 1} \max \set {\norm x^n , 1} }} {{end-eqn}} Taking $n$th roots yields: :$\norm {x + 1_R} \le \paren {n + 1}^{1/n} \max \set {\norm x, 1}$ ==== Lemma 3 ==== {{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3}}{{qed|lemma}} By the Multiple Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} \max \set {\norm x, 1} = \max \set {\norm x, 1}$ By Inequality Rule for Real Sequences: :$\norm {x + 1_R} \le \max \set {\norm x, 1}$ The result follows.
{{begin-eqn}} {{eqn | l = \norm {x + y}       | o = \le       | r = \max \set {\norm x, \norm y} }} {{eqn | ll= \leadstoandfrom       | l = \norm {x + y} \norm {y^{-1} }       | o = \le       | r = \max \set {\norm x \norm {y^{-1} }, \norm y \norm {y^{-1} } }       | c = Multiply through by $\norm{y^{-1} }$ }} {{eqn | ll= \leadstoandfrom       | l = \norm {\paren {x + y} y^{-1} }       | o = \le       | r = \max \set {\norm {x y^{-1} }, \norm {y y^{-1} } }       | c = {{NormAxiom|2}} }} {{eqn | ll= \leadstoandfrom       | l = \norm {\paren {x y^{-1} + y y^{-1} } }       | o = \le       | r = \max \set {\norm {x y^{-1} }, \norm {y y^{-1} } }       | c = Ring axiom $\text D$: Product is Distributive over Addition }} {{eqn | ll= \leadstoandfrom       | l = \norm {\paren {x y^{-1} + 1_R } }       | o = \le       | r = \max \set {\norm {x y^{-1} }, \norm {1_R } }       | c = Product with Inverse is Unit }} {{eqn | ll= \leadstoandfrom       | l = \norm {\paren {x y^{-1} + 1_R } }       | o = \le       | r = \max \set {\norm {x y^{-1} }, 1 }       | c = Norm of Unity }} {{end-eqn}} {{qed}}
By definition of central subgroup: :$Z \subseteq \map Z G$ where $\map Z G$ denotes the center of $G$. From Image under Epimorphism of Center is Subset of Center: :$\theta \sqbrk {\map Z G} \subseteq \map Z H$ From Image of Subset under Mapping is Subset of Image it follows that: :$\theta \sqbrk Z \subseteq \map Z H$ The result follows. {{qed}} Category:Central Subgroups Category:Group Epimorphisms 9fdj8xamuhphxaj2xl5xszjyb657b1r
Let $\tuple {z, w} \in Z \times W$. Let $\tuple {x, y} \in G \times H$. Then: {{begin-eqn}} {{eqn | l = \tuple {x, y} \tuple {z, w}       | r = \tuple {x z, y w}       | c = {{Defof|Group Direct Product}} }} {{eqn | r = \tuple {z x, w y}       | c = {{Defof|Central Subgroup}} }} {{eqn | r = \tuple {z, w} \tuple {x, y}       | c = {{Defof|Group Direct Product}} }} {{end-eqn}} $\tuple {x, y}$ is arbitrary. Thus $\tuple {z, w}$ commutes with all elements of $G \times H$. Hence $\tuple {z, w}$ is in the center of $G \times H$ That is, $Z \times W$ is a subgroup of the center $G \times H$. The result follows by definition of central subgroup. {{qed}} Category:Central Subgroups Category:Group Direct Products t3bozwduh2dv43876c7vq6v7lgk7520
{{AimForCont}} $D_4$ is the internal group product of $2$ proper subgroups $H$ and $K$ of $D_4$. {{WLOG}}, let $\order H = 2$ and $\order K = 4$. $H$ needs to be normal in $D_4$ for the conditions of the internal group product to be satisfied. So the non-identity element of $H$ needs to be conjugated to itself by every element of $G$. This means $H$ is a subset of the center of $G$. From Center of Dihedral Group $D_4$: :$H = \set {e, a^2}$ But every non-trivial normal subgroup of $D_4$ contains $a^2$. This makes it impossible for $H \cap K = \set e$. Thus $D_4$ cannot be an internal group product. {{qed}}
Let $G$ be defined by its group presentation: :$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$ Let $z$ denote $x^5$. Then: {{begin-eqn}} {{eqn | l = y z y^{-1}       | r = y x^5 y^{-1}       | c =  }} {{eqn | r = \paren {y x y^{-1} }^5       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = \paren {x^4}^5       | c = from Group Presentation }} {{eqn | r = x^{20}       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = x^{15} x^5       | c = Powers of Group Elements: Sum of Indices }} {{eqn | r = e \cdot x^5       | c = from Group Presentation }} {{eqn | r = x^5       | c = {{GroupAxiom|2}} }} {{eqn | r = z       | c = Definition of $z$ }} {{eqn | ll= \leadsto       | l = y z       | r = z y       | c = Product of both sides with $y$ }} {{end-eqn}} So $z$ commutes with $y$. As $z$ is a power of $x$, $z$ also commutes with $x$. Hence by definition of center: :$z \in \map Z G$ It follows that $\gen z$ is a normal subgroup of order $3$. Let $N$ be the subgroup of $G$ generated by $x^3$ and $y$. Note that: {{begin-eqn}} {{eqn | l = y x^3 y^{-1}       | r = \paren {y x y^{-1} }^3       | c = Power of Conjugate equals Conjugate of Power }} {{eqn | r = \paren {x^4}^5       | c = from Group Presentation }} {{eqn | r = x^{12}       | c = Powers of Group Elements: Product of Indices }} {{eqn | r = x^{-3}       | c = as $x^{15} = e$ }} {{end-eqn}} Hence the generator of $N$ satisfies: :$\paren{x^3}^5 = e = y^2$ and: :$y x^3 y^{-1} = x^{-3}$ Let $w := x^3$. Then $N$ is generated by $w$ and $y$ where: :$w^5 = 1 = y^2$ and: :$w y = y w^4 = y w^{-1}$ and it is seen that $N$ is isomorphic to the dihedral group $D_5$. It is now to be shown that $G$ is an internal group direct product of $N$ and $\gen z$. We have that $N \cap \gen z = \set e$. From the Internal Direct Product Theorem, we need to prove only that $N$ is a normal subgroup of $G$. We have that: :$\order N = 10$ where $\order N$ denotes the order of $N$. We also have that $N$ is a subgroup of its normalizer $\map {N_G} N$. Hence by Lagrange's Theorem: :$10 \divides \order {\map {N_G} N}$ where $\divides$ denotes divisibility. Again by Lagrange's Theorem: :$\order {\map {N_G} N} \divides 30$ We have: {{begin-eqn}} {{eqn | l = x w x^{-1}       | r = x x^3 x^{-1}       | c = Definition of $w$ }} {{eqn | r = x^3       | c =  }} {{eqn | r = w       | c =  }} {{end-eqn}} demonstrating that $x$ is conjugate to $w$. Then: {{begin-eqn}} {{eqn | l = x y x^{-1}       | r = x y^{-1} x^{-1}       | c = as $y^2 = e$ }} {{eqn | r = y x^4 x^{-1}       | c = from Group Presentation: $y x y^{-1} = x^4$ }} {{eqn | r = y x^3       | c =  }} {{eqn | r = y w       | c =  }} {{eqn | o = \in       | r = N       | c =  }} {{end-eqn}} demonstrating that $x$ is conjugate to $y$. Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 10$. As $10 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 30$, it follows that: :$\map {N_G} N = G$ and so $N$ is normal in $G$. Thus: : $N$ and $\gen z$ are normal in $G$ : $N \cap \gen z = \set e$ : $N \gen z = G$ and it therefore follows from the  Internal Direct Product Theorem that: :$G = C_3 \times D_5$ {{qed}}
{{ProofWanted|This is probably just a statement of Fundamental Theorem of Finite Abelian Groups, which needs to be studied to see what it actually means}}
{{begin-eqn}} {{eqn | l = \overline {T_1 \cap T_2}       | r = \mathbb U \setminus \paren {T_1 \cap T_2}       | c = {{Defof|Set Complement}} }} {{eqn | r = \paren {\mathbb U \setminus T_1} \cup \paren {\mathbb U \setminus T_2}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \overline {T_1} \cup \overline {T_2}       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in \overline {T_1 \cap T_2} }} {{eqn | o = \leadstoandfrom       | r = x \notin \paren {T_1 \cap T_2}       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {x \in T_1 \land x \in T_2}       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {x \in T_1} \lor \neg \paren {x \in T_2}       | c = De Morgan's Laws (Logic): Disjunction of Negations }} {{eqn | o = \leadstoandfrom       | r = x \in \overline {T_1} \lor x \in \overline {T_2}       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadstoandfrom       | r = x \in \overline {T_1} \cup \overline {T_2}       | c =  }} {{end-eqn}} By definition of set equality: :$\overline {T_1 \cap T_2} = \overline {T_1} \cup \overline {T_2}$ {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === {{:Parallelism is Equivalence Relation/Reflexivity}}{{qed|lemma}} === Symmetry === {{:Parallelism is Equivalence Relation/Symmetry}}{{qed|lemma}} === Transitivity === {{:Parallelism is Equivalence Relation/Transitivity}}{{qed|lemma}} $\parallel$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}