hcju/proofwikiqa huggingface dataset:
Let the distinct orbits of $G$ under the conjugacy action be: :$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$ Then from the Partition Equation: :$\order G = \order {\Orb {x_1} } + \order {\Orb {x_2} } + \cdots + \order {\Orb {x_s} }$ From the Orbit-Stabilizer Theorem: :$\order {\Orb {x_i} } \divides \order G, i = 1, \ldots, s$ The result follows from the definition of the conjugacy action. {{Qed}}
Let $G$ be a group of order $p^2$, where $p$ is prime. Let $\map Z G$ be the center of $G$. By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$. By Lagrange's Theorem: :$\order {\map Z G} \divides \order G$ It follows that $\order {\map Z G} = 1, p$ or $p^2$. By Center of Group of Prime Power Order is Non-Trivial: :$\order {\map Z G} \ne 1$ Suppose $\order {\map Z G} = p$. Then: {{begin-eqn}} {{eqn | l = \order {\map Z G}       | r = \index G {\map Z G}       | c = {{Defof|Index of Subgroup}} }} {{eqn | r = \order G / \order {\map Z G}       | c = Lagrange's Theorem }} {{eqn | r = p^2 / p       | c =  }} {{eqn | r = p       | c =  }} {{end-eqn}} So $G / \map Z G$ is non-trivial, and of prime order. From Prime Group is Cyclic, $G / \map Z G$ is a cyclic group. But by Quotient of Group by Center Cyclic implies Abelian, that cannot be the case. Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$. Therefore $G$ is abelian. {{qed}}
Suppose $G$ is abelian. From Group equals Center iff Abelian: : $\map Z G = G$ and the result is seen to be true as $G$ is itself non-trivial. From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$. So, suppose $G$ is non-abelian. Thus $\map Z G \ne G$ and therefore $G \setminus \map Z G \ne \O$. Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $G \setminus \map Z G$ is partitioned. From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element. From the Conjugacy Class Equation: :$\displaystyle \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$ From Number of Conjugates is Number of Cosets of Centralizer: :$\order {\conjclass {x_j} } \divides \order G$ Let $\map {N_G} x$ be the normalizer of $x$ in $G$. Then: :$\forall j: 1 \le j \le m: \index G {\map {N_G} {x_j} } > 1 \implies p \divides \index G {\map {N_G} {x_j} }$ Since $p \divides \order G$, it follows that: : $p \divides \order {\map Z G}$ and the result follows. {{Qed}}
{{AimForCont}} $\order {\map Z G} = p^2$. Then $\order {G / \map Z G} = p$. From Prime Group is Cyclic it follows that $G / \map Z G$ is cyclic. From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian. From Group equals Center iff Abelian it follows that $\order {\map Z G} = p^3$. The result follows by Proof by Contradiction. {{qed}} {{expand|This results generalizes naturally to a group of order $p^n$: its center is not of order $p^{n-1}$.}} Category:Normal Subgroups Category:Centers of Groups Category:P-Groups 37p385xpkzyb8mfwohstvgjm9u679z5
From Center of Group is Normal Subgroup, $\map Z G \lhd G$. By Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial. If $\map Z G$ is a proper subgroup, the proof is finished. Otherwise, $\map Z G = G$. Then $G$ is abelian by Group equals Center iff Abelian. However, then any $a \in G: a \ne e$ generates a non-trivial normal subgroup $\gen a$, as Subgroup of Abelian Group is Normal. If $\order a = \order {\gen a} < \order G$, the proof is complete. Otherwise: :$\order a = \order G = p^n$ Then $\order {a^p} = p^{n - 1}$ from Subgroup of Finite Cyclic Group is Determined by Order, so $a^p$ generates that non-trivial normal subgroup. {{qed}}
To be proved by induction on $n$. Let $P_n$ be the proposition for $\order G = p^n$. === Basis for the Induction === $P_1$ is trivially true because: :$\set e = G_0 \subset G_1 = G$ From Prime Group is Cyclic, a group whose order is prime is cyclic. === Induction Hypothesis === Suppose $P_k$ is true for all groups of order $p^k$ for all $k < n$. We need to show that this implies $P_{k + 1}$ is true. === Induction Step === Let $G$ be a group of order $p^n$. By Prime Power Group has Non-Trivial Proper Normal Subgroup, $G$ has a proper non-trivial normal subgroup. There will be a finite number of these, so we are free to pick one of maximal order. We call this $H$, such that $\order H = p^t, t < n$. We need to show that $t = n - 1$. Suppose $t < n - 1$. Then $G / H$ is a group of order $p^{n - t} \ge p^2$. Again by Prime Power Group has Non-Trivial Proper Normal Subgroup, $G / H$ has a proper non-trivial normal subgroup, which we will call $N$. Let $H' = \set {g \in G: g H \in N}$. We now show that $H \lhd G$. Let $g, g' \in H'$. Then $g H, g' H \in N$. Since $N < G / H$: : $\paren {g H} \paren {g' H} = g g' H \in N$ and so $g g' \in N$. If $g \in H$, then $g H \in N$. Since $N < G / H$: :$\paren {g H}^{-1} = g^{-1} H \in N$ and so $g^{-1} \in H'$. Next: {{begin-eqn}} {{eqn | l = \paren {H'}^a       | r = \set {g \in G: a g a^{-1} \in H'}       | c =  }} {{eqn | r = \set {g \in G: a g a^{-1} H \in N}       | c =  }} {{eqn | r = \set {g \in G: \paren {a H} \paren {g H} \paren {a H}^{-1} \in N}       | c =  }} {{eqn | r = \set {g \in G: \paren {g H} \in N^{a H} }       | c = {{Defof|Conjugate of Group Subset}} }} {{eqn | r = \set {g \in G: \paren {g H} \in N}       | c = as $N$ is normal }} {{eqn | r = H'       | c =  }} {{end-eqn}} So clearly $H' / H = N$, therefore: :$\dfrac {\order {H'} } {\order H} = \index {H'} N = \order N \ge p$ So: :$\order {H'} \ge p \order H$ contradicting the maximality of $\order H$. It follows that $t = n - 1$. Finally, we set $G_{n - 1} = H$ and use induction to show that $P_{n - 1}$ holds. Since $G / H = G / G_{n - 1}$ is a group of order $p$, then it is automatically cyclic. {{qed}}
{{handwaving}} The validity of this rule follows directly from the definition of multiplication of integers. The product $a b$ (for $a, b \in \N_{>0}$) is the number of sequences $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any one of $b$ items. {{qed}}
The mapping $\theta: G \times H \to H \times G$ defined as: :$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$ is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows: === Injective === {{begin-eqn}} {{eqn | l = \map \theta {g_1, h_1}       | r = \map \theta {g_2, h_2}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {h_1, g_1}       | r = \tuple {h_2, g_2}       | c = Definition of $\theta$ }} {{eqn | ll= \leadsto       | l = \tuple {g_1, h_1}       | r = \tuple {g_2, h_2}       | c = Equality of Ordered Pairs }} {{end-eqn}} Thus $\theta$ is injective by definition. {{qed|lemma}} === Surjective === {{begin-eqn}} {{eqn | l = \tuple {h, g}       | o = \in       | r = H \times G       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {g, h}       | o = \in       | r = G \times H       | c =  }} {{eqn | ll= \leadsto       | lo= \exists \tuple {g, h} \in G \times H:       | l = \tuple {h, g}       | r = \map \theta {g, h}       | c =  }} {{end-eqn}} Thus $\theta$ is surjective by definition. {{qed|lemma}} === Group Homomorphism === Let $\tuple {g_1, h_1}, \tuple {g_2, h_2} \in G \times H$. Then: {{begin-eqn}} {{eqn | l = \map \theta {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} }       | r = \map \theta {g_1 \circ_g g_2, h_1 \circ_h h_2}       | c = Definition of $\circ$ }} {{eqn | r = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}       | c = Definition of $\theta$ }} {{eqn | r = \tuple {h_1, g_1} \star \tuple {h_2, g_2}       | c = Definition of $\star$ }} {{eqn | r = \map \theta {g_1, h_1} \star \map \theta {g_2, h_2}       | c = Definition of $\theta$ }} {{end-eqn}} thus proving that $\theta$ is a homomorphism. {{Qed}}
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups whose identities are $e_G$ and $e_H$ respectively. Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both abelian. Then from External Direct Product Commutativity, $\struct {G \times H, \circ}$ is also abelian. Now suppose that $\struct {G \times H, \circ}$ is abelian. Then: {{begin-eqn}} {{eqn | l = \tuple {g_1 \circ_1 g_2, e_H}       | r = \tuple {g_1 \circ_1 g_2, e_H \circ_2 e_H}       | c = Definition of $e_H$ }} {{eqn | r = \tuple {g_1, e_H} \circ \tuple {g_2, e_H}       | c = {{Defof|Group Direct Product}} }} {{eqn | r = \tuple {g_2, e_H} \circ \tuple {g_1, e_H}       | c = as $\struct {G \times H, \circ}$ is abelian }} {{eqn | r = \tuple {g_2 \circ_1 g_1, e_H \circ_2 e_H}       | c = {{Defof|Group Direct Product}} }} {{eqn | r = \tuple {g_2 \circ_1 g_1, e_H}       | c = Definition of $e_H$ }} {{end-eqn}} Thus: : $g_1 \circ_1 g_2 = g_2 \circ_1 g_1$ and $\struct {G, \circ_1}$ is seen to be abelian. The same argument holds for $\struct {H, \circ_2}$. {{qed}}
Let $G, H, K$ be groups. The mapping $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as: :$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$ is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows: === Injective === Let $\map \theta {\tuple {g_1, \tuple {h_1, k_1} } } = \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }$. By the definition of $\theta$: :$\tuple {\tuple {g_1, h_1}, k_1} = \tuple {\tuple {g_2, h_2}, k_2}$ By Equality of Ordered Pairs: :$\tuple {g_1, h_1} = \tuple {g_2, h_2}$ :$k_1 = k_2$ and consequently: :$g_1 = g_2, h_1 = h_2$ Thus: :$\tuple {g_1, \tuple {h_1, k_1} } = \tuple {g_2, \tuple {h_2, k_2} }$ and so $\theta$ is seen to be injective. === Surjective === If $\tuple {\tuple {g, h}, k} \in \paren {G \times H} \times K$, then $g \in G, h \in H, k \in K$. Thus: :$\tuple {g, \tuple {h, k} } = \in G \times \paren {H \times K}$ and: :$\map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$ and so $\theta$ is seen to be surjective. === Group Homomorphism === Now let $\tuple {g_1, \tuple {h_1, k_1} }, \tuple {g_2, \tuple {h_2, k_2} } \in G \times \tuple {H \times K}$. Then: {{begin-eqn}} {{eqn | l = \map \theta {\tuple {g_1, \tuple {h_1, k_1} } \tuple {g_2, \tuple {h_2, k_2} } }       | r = \map \theta {\tuple {g_1 g_2, \tuple {h_1, k_1} \tuple {h_2, k_2} } }       | c =  }} {{eqn | r = \map \theta {\tuple {g_1 g_2, \tuple {h_1 h_2, k_1 k_2} } }       | c =  }} {{eqn | r = \tuple {\tuple {g_1 g_2, h_1 h_2}, k_1 k_2}       | c =  }} {{eqn | r = \tuple {\tuple {g_1, h_1} \tuple {g_2, h_2}, k_1 k_2}       | c =  }} {{eqn | r = \tuple {\tuple {g_1, h_1}, k_1} \tuple {\tuple {g_2, h_2}, k_2}       | c =  }} {{eqn | r = \map \theta {\tuple {g_1, \tuple {h_1, k_1} } } \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }       | c =  }} {{end-eqn}} showing that $\theta$ is a (group) homomorphism. The result follows. {{qed}} Category:Group Direct Products Category:Associativity 0e07ovfdbn3d1ksqoxy8c7rugsa4xz5
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively. === Necessary condition === Suppose: : $(1): \quad \order G = n, G = \gen x$ : $(2): \quad \order H = m, H = \gen y$ : $(3): \quad m \perp n$ Then: {{begin-eqn}} {{eqn | l = \order {\tuple {x, y} }       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {x, y}^k       | r = e_{G \times H} = \tuple {x^k, y^k}       | c = Powers of Elements in Group Direct Product }} {{eqn | ll= \leadsto       | l = x^k       | r = e_G, y^k = e_H       | c =  }} {{eqn | ll= \leadsto       | l = n       | o = \divides       | r = k, m \divides k       | c =  }} {{eqn | ll= \leadsto       | l = \lcm \set {n, m}       | o = \divides       | r = k       | c =  }} {{eqn | ll= \leadsto       | l = n m       | o = \divides       | r = k       | c = as $n \perp m$ }} {{end-eqn}} But then: :$\tuple {x, y}^{n m} = e_{G \times H} = \tuple {x^{n m}, y^{n m} }$ Thus: :$k \divides n m$ So: :$\order {\tuple {x, y} } = n m \implies \gen {\tuple {x, y} } = G \times H$ {{qed|lemma}} === Sufficient condition === Suppose that $G \times H$ is cyclic. Let $\tuple {x, y}$ be a generator of $G \times H$. By Cardinality of Cartesian Product the order of $G \times H$ is: :$\order G \cdot \order H = g h$ Therefore by Order of Cyclic Group equals Order of Generator: :$\order {\tuple {x, y} } = g h$ On the other hand, by Order of Group Element in Group Direct Product we have: :$\order {\tuple {x, y} } = \lcm \set {\tuple {\order x, \order y} }$ Next we claim that $x$ generates $G$. Let $x' \in G$. Then: :$\tuple {x', e_H} \in G \times H$ so there exists $k \in \N$ such that: :$\tuple {x, y}^k = \tuple {x^k, y^k} = \tuple {x', e_H}$ and therefore $x^k = x'$. Thus the powers of $x$ generate the whole group $G$. In the same way, it is seen that $y$ generates $H$. Therefore by Order of Cyclic Group equals Order of Generator: :$\order x = g$ :$\order y = h$ Thus we have that: :$g h = \order {\tuple {x, y} } = \lcm \set {g, h}$ Moreover by Product of GCD and LCM we have that: :$\lcm \set {g, h} = \dfrac {g h} {\gcd \set {g, h} }$ These two equalities imply that: :$\gcd \set {g, h} = 1$ That is, $g$ and $h$ are coprime. {{qed}}
Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be infinite cyclic groups. Let $G = \struct {G, \circ} = G_1 \times G_2$. Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$. From Generators of Infinite Cyclic Group: :$g_1$ and $g_1^{-1}$ are the only generators of $G_1$ :$g_2$ and $g_2^{-1}$ are the only generators of $G_2$. So a generator of $G$ must be of the form $\tuple { g_1^{\pm 1}, g_2^{\pm 1} }$ to have any hope to generate all of $G$. {{WLOG}}, suppose that $G = \gen {\tuple {g_1, g_2} }$. Let $e_1$ be the identity element of $G_1$. Let $x_2 \in G_2$. From the definition of an infinite cyclic group, both $g_1$ and $g_2$ are of infinite order. Suppose now $\tuple {e_1, x_2} \in \gen {\tuple {g_1, g_2} }$. Then there is an $i \in \Z$ such that we have: :$\tuple {g_1, g_2}^i = \tuple {g_1^i, g_2^i} = \tuple {e_1, x_2}$ However, as $g_1$ is of infinite order, it must be that $i = 0$.  Hence $x_2 = g_2^0 = e_2$, where $e_2$ is the identity element of $G_2$. It follows that $\tuple {e_1, g_2} \notin \gen {\tuple {g_1, g_2} }$. Therefore: :$G \ne \gen {\tuple {g_1, g_2} }$ It follows that $G$ cannot be generated by one element. {{qed}}
Let $G$ and $H$ be a groups whose identities are $e_G$ and $e_H$. Let $l = \lcm \set {m, n}$. From the definition of lowest common multiple: :$\exists x, y \in \Z: l = m x = n y$ From the definition of order of an element: :$g^m = e_G, h^n = e_H$ Thus: {{begin-eqn}} {{eqn | l = \tuple {g, h}^l       | r = \tuple {g^l, h^l}       | c =  }} {{eqn | r = \tuple {\paren {g^m}^x, \paren {h^n}^y}       | c =  }} {{eqn | r = \tuple {e_G^x, e_H^y}       | c =  }} {{eqn | r = \tuple {e_G, e_H}       | c =  }} {{end-eqn}} Now suppose $\exists k \in \Z: \tuple {g, h}^k = \tuple {e_G, e_H}$. It follows that: :$\tuple {g, h}^k = \tuple {g^k, h^k} = \tuple {e_G, e_H}$ Thus: :$g^k = e_G, h^k = e_H$ It follows from $g^k = e_G$ and $\order g = m$, by Absolute Value of Integer is not less than Divisors, that: :$m \divides k$ Similarly it follows that: :$n \divides k$ So $k$ is a positive common multiple of both $m$ and $n$. Since $l$ is the lowest common multiple: :$l \le k$ Therefore $l$ is the smallest such that $\tuple {g, h}^l = \tuple {e_G, e_H}$. The result follows by definition of the order of an element. {{qed}} Category:Group Direct Products qn3x5oeiex39ck218cbpsj3tjzn6b74
=== Necessary Condition === Let $\phi$ be a surjection. Then $\Img \phi$ consists of all the products $\displaystyle \prod_{k \mathop = 1}^n h_k$ such that: :$\forall k \in \closedint 1 n: h_k \in H_k$ Thus, as $\phi$ is surjective, every element of $G$ must be representable in this form. Using the notation: :$\displaystyle \prod_{k \mathop = 1}^n H_k = \set {\prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k}$ the result follows immediately. Thus every element of $G$ is the product of an element of each of $H_k$. {{qed|lemma}} === Sufficient Condition === Let $\displaystyle G = \prod_{k \mathop = 1}^n H_k$. If this is the case, then $g$ can be written as: :$\displaystyle g = \prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k$ Thus: :$g = \map \phi {h_1, h_2, \ldots, h_n}$ and $\phi$ is surjective. {{qed}} Category:Internal Group Direct Products 19kysqrglzqel7vkr389ikpxd1jxabb
=== Necessary Condition === Let $\phi$ be an injection. Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$. As $\phi$ is injective, this means that: :$\tuple {h_1, h_2} = \tuple {k_1, k_2}$ and thus: :$h_1 = k_1, h_2 = k_2$ From the definition of $\phi$, this means: :$h_1 h_2 = k_1 k_2$ Thus, each element of $G$ that can be expressed as a product of the form $h_1 h_2$ can be thus expressed ''uniquely''. Now, suppose $h \in H_1 \cap H_2$. We have: {{begin-eqn}} {{eqn | l = h = h e       | o = :       | r = h \in H_1, e \in H_2       | c =  }} {{eqn | l = h = e h       | o = :       | r = e \in H_1, h \in H_2       | c =  }} {{end-eqn}} Thus we see that: {{begin-eqn}} {{eqn | l = \map \phi {h, e}       | r = \map \phi {e, h}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {h, e}       | r = \tuple {e, h}       | c =  }} {{eqn | ll= \leadsto       | l = h       | r = e       | c =  }} {{end-eqn}} Thus: :$H_1 \cap H_2 = \set e$ {{qed|lemma}} === Sufficient Condition === Let $H_1 \cap H_2 = \set e$. Let: :$\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$ Then: :$h_1 h_2 = k_1 k_2: h_1, k_1 \in H_1, h_2, k_2 \in H_2$ Thus: :$k_1^{-1} h_1 = k_2 h_2^{-1}$ But: :$k_1^{-1} h_1 \in H_1$ and $k_2 h_2^{-1} \in H_2$ As they are equal, we have: :$k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \set e$ It follows that: :$h_1 = k_1, h_2 = k_2$ and thus: :$\tuple {h_1, h_2} = \tuple {k_1, k_2}$ Thus $\phi$ is injective and the result follows. {{qed}} Category:Internal Group Direct Products Category:Injections 8dkzl8zwmhesfwrzsqdn1xfvrontyp8
$\phi$ is an isomorphism, so in particular a (group) homomorphism. Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$. Now suppose $a \in G$. As $\phi$ is an isomorphism, it follows that $\phi$ is surjective. Thus by Subgroup Product is Internal Group Direct Product iff Surjective: :$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$ Now any element of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$. Thus: {{begin-eqn}} {{eqn | l = a h a^{-1}       | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}       | c =  }} {{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}       | c =  }} {{eqn | r = h_1 h h_2 h_2^{-1} h_1^{-1}       | c = $h \in H_1$ and $h_2 \in H_2$ commute }} {{eqn | r = h_1 h h_1^{-1} \in H_1       | c =  }} {{end-eqn}} Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a normal subgroup of $G$. Similarly, any element of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$. Thus: {{begin-eqn}} {{eqn | l = a h a^{-1}       | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}       | c =  }} {{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}       | c =  }} {{eqn | r = h_1 h' h_1^{-1}       | c = as $h' = h_2 h h_2^{-1} \in H_2$ }} {{eqn | r = h' h_1 h_1^{-1}       | c = $h_1 \in H_1$ and $h' \in H_2$ commute }} {{eqn | r = h' \in H_2       | c =  }} {{end-eqn}} Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well. {{qed}} Category:Internal Group Direct Products Category:Group Isomorphisms Category:Normal Subgroups tal7o18xi8em9ebt61g1ghj7qd3spdp
Let $H_1$ and $H_2$ be normal subgroups of $G$. Let $h_1 \in H_1, h_2 \in H_2$. Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$. {{begin-eqn}} {{eqn | l = x       | r = h_1 h_2 h_1^{-1} h_2^{-1}       | c =  }} {{eqn | r = \paren {h_1 h_2 h_1^{-1} } h_2^{-1}       | c =  }} {{end-eqn}} As $H_2$ is normal, we have $h_1 h_2 h_1^{-1} \in H_2$ and thus $x \in H_2$. Similarly, we can show that $x \in H_1$ and so $x \in H_1 \cap H_2$ and thus $x = e$. From Product of Commuting Elements with Inverses, $h_1 h_2 h_1^{-1} h_2^{-1} = e$ {{iff}} $h_1$ and $h_2$ commute. Thus $h_1 h_2 = h_2 h_1$.  As $h_1$ and $h_2$ are arbitrary elements of $H_1$ and $H_2$, it follows that every element of $H_1$ commutes with every element of $H_2$. Thus from Induced Group Product is Homomorphism iff Commutative, $\phi$ is a homomorphism. {{qed}} Category:Internal Group Direct Products Category:Normal Subgroups Category:Group Homomorphisms dr1zduk92t4fsu9xcfdmjshzwap7ogp
=== Sufficient Condition === Let $G$ be the internal group direct product of $H_1$ and $H_2$. :$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 \circ H_2$. :$(2): \quad$ From Internal Group Direct Product is Injective, $H_1$ and $H_2$ are independent subgroups of $G$. :$(3): \quad$ From Internal Group Direct Product Isomorphism, $H_1 \lhd G$ and $H_2 \lhd G$. {{qed|lemma}} === Necessary Condition === Let $C: H_1 \times H_2 \to G$ be the mapping defined as: :$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map C {h_1, h_2} = h_1 \circ h_2$ Suppose the three conditions hold. :$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective. :$(2): \quad$ From Internal Group Direct Product is Injective, $C$ is injective. :$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism. Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism. So by definition, $G$ is the internal group direct product of $H_1$ and $H_2$. {{qed}}
It is apparent from the definitions of both the inclusion mapping and the identity mapping that: :$(1): \quad \Dom {i_S} = S = \Dom {I_S}$ :$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$ === Necessary Condition === Let $i_S: S \to T = I_S: S \to S$. From Equality of Mappings, we have that the codomain of $i_S$ equals the codomain of $I_S$. Thus the codomain of $i_S$ equals the codomain of $I_S$ equals $S$ and thus $T = S$. So $\forall s \in S: s = \map {i_S} s$ and so $i_S$ is surjective. {{qed|lemma}} === Sufficient Condition === Now let $i_S: S \to T$ be a surjection. Then: :$\forall s \in T: s = \map {i_S} s \implies s \in S$ and therefore: :$T \subseteq S$ Thus: :$T = S$ and so the codomain of $i_S$ equals the codomain of $I_S$ which equals $S$. Thus $i_S: S \to T = I_S: S \to S$. {{qed}}
From Group Direct Product of Cyclic Groups noting that $2 \perp 3$: :$C_2 \times C_3 \cong C_6$ * $C_6$ is the internal group direct product of $C_2$ and $C_3$: Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left \langle {x^3} \right \rangle$, and let $\left({C_3, \circ}\right) = \left \langle {x^2} \right \rangle$. From Subgroup of Abelian Group is Normal, $C_2 \lhd C_6$ and $C_3 \lhd C_6$. We can factorise the elements of $C_6$ thus: :$e = e \circ e, x = x^3 \circ \left({x^2}\right)^2, x^2 = e \circ x^2, x^3 = x^3 \circ e, x^4 = e \circ \left({x^2}\right)^2, x^5 = x^2 \circ x^3$  Hence the result. {{qed}} Category:Examples of Cyclic Groups Category:Groups of Order 6 krip2kmx54xcemcyjmzixdpxw4ne1rd