hcju/proofwikiqa huggingface dataset:
Let $k$ be the order of $x$ in $G$ according to Definition 1. Let $l$ be the order of $x$ in $G$ according to Definition 3. === Definition $1$ is equivalent to Definition $3$ === {{AimForCont}} $k \ne l$. Then from Ordering on 1-Based Natural Numbers is Trichotomy either $k < l$ or $k > l$. According to Definition 3: :$\forall i, j \in \Z: 0 \le i < j < l \implies x^i \ne x^j$ If $k < l$, then letting $i = 0$ and $j = k$ yields a contradiction by Element to Power of Zero is Identity. If $k > l$, then: :$\exists i,j \in \Z: 0 \le i < j < k: x^i = x^j$ But then $x^{j - i} = e$ and $j - i < k$, contradicting the fact that $k \in \Z_{> 0}$ is the smallest such that $x^k = e$. Therefore $k = l$. {{qed|lemma}} === Definition $1$ is equivalent to Definition $2$ === It follows straight away from List of Elements in Finite Cyclic Group that $\order {\gen a} = k$: :$\gen a = \set {a^0, a^1, a^2, \ldots, a^{k - 1} }$ {{qed}}
Proof by induction: Let $n \in \Z$. For all $m \in \N$, let $\map P m$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$ $\map P 0$ is true, as this just says: :$\dbinom n n = \dbinom {n + 1} {n + 1}$ But $\dbinom n n = \dbinom {n + 1} {n + 1} = 1$ from the {{Defof|Binomial Coefficient}}. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^1 \binom {n + j} n       | r = \binom n n + \binom {n + 1} n       | c =  }} {{eqn | r = 1 + \paren {n + 1}       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = n + 2       | c =  }} {{eqn | r = \binom {n + 2} {n + 1}       | c = {{Defof|Binomial Coefficient}} }} {{end-eqn}} So: :$\displaystyle \sum_{j \mathop = 0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$ and $\map P 1$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k \binom {n + j} n = \binom {n + k + 1} {n + 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 0}^{k+1} \binom {n + j} n = \binom {n + k + 2} {n + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{k + 1} \binom {n + j} n       | r = \sum_{j \mathop = 0}^k \binom {n + j} n + \binom {n + k + 1} n       | c =  }} {{eqn | r = \binom {n + k + 1} {n + 1} + \binom {n + k + 1} n       | c = Induction Hypothesis }} {{eqn | r = \binom {n + k + 2} {n + 1}       | c = Pascal's Rule }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$ Finally, from Symmetry Rule for Binomial Coefficients: :$\dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$ {{qed}}
From Sum of Binomial Coefficients over Lower Index we have: :$\displaystyle \sum_{i \mathop \in \Z} \binom n i = 2^n$ That is: :$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \cdots + \dbinom n n = 2^n$ as $\dbinom n i = 0$ for $i < 0$ and $i > n$. This can be written more conveniently as: :$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$ Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have: :$\displaystyle \sum_{i \mathop \in \Z} \left({-1}\right)^i \binom n i = 0$ That is: :$\dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$ Adding them together, we get: :$2 \dbinom n 0 + 2 \dbinom n 2 + 2 \dbinom n 4 + \cdots = 2^n$ as the odd index coefficients cancel out. Dividing by $2$ throughout gives us the result. {{qed}}
From Sum of Binomial Coefficients over Lower Index we have: :$\displaystyle \sum_{i \mathop \in \Z} \binom n i = 2^n$ That is: :$\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \cdots + \dbinom n n = 2^n$ as $\dbinom n i = 0$ for $i < 0$ and $i > n$. This can be written more conveniently as: :$(1): \quad \dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 + \dbinom n 4 + \cdots = 2^n$ Similarly, from Alternating Sum and Difference of Binomial Coefficients for Given n we have: :$\displaystyle \sum_{i \mathop \in \Z} \paren {-1}^i \binom n i = 0$ That is: :$(2): \quad \dbinom n 0 - \dbinom n 1 + \dbinom n 2 - \dbinom n 3 + \dbinom n 4 - \cdots = 0$ Subtracting $(2)$ from $(1)$, we get: :$2 \dbinom n 1 + 2 \dbinom n 3 + 2 \dbinom n 5 + \cdots = 2^n$ as the even index coefficients cancel out. Dividing by $2$ throughout gives us the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^n j \binom n j       | r = \sum_{j \mathop = 1}^n j \binom n j       | c = as $0 \dbinom n 0 = 0$ }} {{eqn | r = \sum_{j \mathop = 1}^n n \binom {n - 1} {j - 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = n \sum_{j \mathop = 0}^{n - 1} \binom {n - 1} j       | c = Translation of Index Variable of Summation }} {{eqn | r = n 2^{n - 1}       | c = Sum of Binomial Coefficients over Lower Index }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^n \left({-1}\right)^{n + 1} j \binom n j       | r = \sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} j \binom n j       | c = as $0 \dbinom n 0 = 0$ }} {{eqn | r = \sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} n \binom {n - 1} {j - 1}       | c = Factors of Binomial Coefficient }} {{eqn | r = n \sum_{j \mathop = 0}^{n - 1} \left({-1}\right)^{n - 1} \binom {n - 1} j       | c = Translation of Index Variable of Summation, and $\left({-1}\right)^{n + 1} = \left({-1}\right)^{n - 1}$ }} {{eqn | r = 0       | c = Alternating Sum and Difference of Binomial Coefficients for Given n }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | ll= \forall x \in \R:       | l = 0       | o = \le       | r = \paren {x \map f t + \map g t}^2 }} {{eqn | l = 0       | o = \le       | r = \int_a^b \paren {x \map f t + \map g t}^2 \rd t       | c = Relative Sizes of Definite Integrals }} {{eqn | r = x^2 \int_a^b \paren {\map f t}^2 \rd t + 2 x \int_a^b \map f t \, \map g t \rd t + \int_a^b \paren {\map g t}^2 \rd t       | c = Linear Combination of Integrals }} {{eqn | r = A x^2 + 2 B x + C       | c =  }} {{end-eqn}} where: {{begin-eqn}} {{eqn | l = A       | r = \int_a^b \paren {\map f t}^2 \rd t }} {{eqn | l = B       | r = \int_a^b \map f t \map g t \rd t }} {{eqn | l = C       | r = \int_a^b \paren {\map g t}^2 \rd t }} {{end-eqn}} The quadratic equation $A x^2 + 2 B x + C$ is non-negative for all $x$. It follows (using the same reasoning as in Cauchy's Inequality) that the discriminant $\paren {2 B}^2 - 4 A C$ of this polynomial must be non-positive. Thus: :$B^2 \le A C$ and hence the result. {{qed}}
Let $w_1, w_2, \ldots, w_n$ and $z_1, z_2, \ldots, z_n$ be arbitrary complex numbers. Take the Binet-Cauchy Identity: :$\displaystyle \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$ and set $a_i = w_i, b_j = \overline {z_j}, c_i = \overline {w_i}, d_j = z_j $. This gives us: {{begin-eqn}} {{eqn | l = \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}       | r = \paren {\sum_{i \mathop = 1}^n w_i z_i} \paren {\sum_{j \mathop = 1}^n \overline {z_j} \overline {w_j} }       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \paren {\overline {w_i} z_j - \overline {w_j} z_i}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\sum_{i \mathop = 1}^n w_i \overline {w_i} } \paren {\sum_{j \mathop = 1}^n \overline {z_j} z_j}       | r = \paren {\sum_{i \mathop = 1}^n w_i z_i} \overline {\paren {\sum_{i \mathop = 1}^n w_i z_i} }       | c =  }} {{eqn | o =        | ro= +       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {w_i \overline {z_j} - w_j \overline {z_i} } \overline {\paren {w_i \overline {z_j} - w_j \overline {z_i} } }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \cmod {z_j}^2}       | r = \cmod {\sum_{i \mathop = 1}^n w_i z_i}^2 + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \cmod {w_i \overline {z_j} - w_j \overline {z_i} }^2       | c = Modulus in Terms of Conjugate }} {{eqn | ll= \leadsto       | l = \paren {\sum_{i \mathop = 1}^n \cmod {w_i}^2} \paren {\sum_{j \mathop = 1}^n \paren {z_j}^2}       | o = \ge       | r = \paren {\sum_{i \mathop = 1}^n w_i z_i}^2       | c = as $\cmod {w_i \overline {z_j} - w_j \overline {z_i} }^2$ cannot be negative }} {{end-eqn}} Hence the result. {{qed}} {{Namedfor|Augustin Louis Cauchy|name2 = Karl Hermann Amandus Schwarz}}
Let $y = \left \lfloor{x}\right \rfloor$. By Floor Function is Integer, we have that $y \in \Z$. Then from Real Number is Integer iff equals Ceiling, we have: :$\left \lceil{y}\right \rceil = y$ So: :$\left \lceil {\left \lfloor {x}\right \rfloor}\right \rceil = \left \lfloor {x}\right \rfloor$ {{qed}}
Let $y = \left \lceil{x}\right \rceil$. By Ceiling Function is Integer, we have that $y \in \Z$. From Real Number is Integer iff equals Floor, we have: :$\left \lfloor{y} \right \rfloor = y$ So: :$\left \lfloor {\left \lceil {x}\right \rceil}\right \rfloor = \left \lceil {x}\right \rceil$ {{qed}}
From Real Number is Integer iff equals Floor: :$x \in \Z \implies x = \left \lfloor {x}\right \rfloor$ From Real Number is Integer iff equals Ceiling: :$x \in \Z \implies x = \left \lceil {x}\right \rceil$ So: :$x \in \Z \implies \left \lfloor {x}\right \rfloor = \left \lceil {x}\right \rceil$ Now let $x \notin \Z$. From the definition of the floor function: :$\left \lfloor {x} \right \rfloor = \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$ From the definition of the ceiling function: :$\left \lceil {x} \right \rceil = \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$ Thus: :$\left \lfloor {x} \right \rfloor < x < \left \lceil {x} \right \rceil$ Hence the result, from the definition of $\inf$ and $\sup$. {{qed}}
From Integer equals Floor iff between Number and One Less we have: :$x - 1 < \floor x \le x$ and so, by multiplying both sides by -1: :$-x + 1 > -\floor x \ge -x$ From Integer equals Ceiling iff between Number and One More we have: :$\ceiling x = n \iff x \le n < x + 1$ Hence: :$-x \le -\floor x < -x + 1 \implies \ceiling {-x} = -\floor x$ {{Qed}}
From Integer equals Ceiling iff between Number and One More we have: :$x \le \left \lceil{x}\right \rceil < x + 1$ and so, by multiplying by -1: :$-x \ge -\left \lceil{x}\right \rceil > -x - 1$ From Integer equals Floor iff between Number and One Less we have: :$\left \lfloor{x}\right \rfloor = n \iff x - 1 < n \le x$ Hence: :$-x - 1 < -\left \lceil{x}\right \rceil \le -x \implies \left \lfloor{-x}\right \rfloor = -\left \lceil{x}\right \rceil$ {{qed}}
Follows from the definitions of coprime and greatest common divisor as follows. When $n \in \Z: n \ne 0$ we have: :$\gcd \left\{{n, 1}\right\} = 1$ Then by definition again: :$\gcd \left\{{n, 0}\right\} = n$ and so when $n = 1$ we have $\gcd \left\{{1, 0}\right\} = 1$. {{qed}}
From the definition of coprime, we have: :$m \perp n \iff \gcd \set {m, n} = 1$ From the definition of greatest common divisor: :$\gcd \set {n, 0} = \size n$ where $\size n$ is the absolute value of $n$. Let $n \in \set {1, -1}$. Then: :$\gcd \set {n, 0} = \size n = 1$ and so $n \perp 0$. Now suppose $n \notin \set {1, -1}$. If $n = 0$ then $\gcd \set {n, 0}$ is not defined. If $n \ne 0$ then: :$\gcd \set {n, 0} = \size n \ne 1$ {{qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$\displaystyle \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$ === Basis for the Induction === $\map P 0$ says: :$\dbinom 0 m = \dbinom 1 {m + 1}$ When $m = 0$ we have by definition: :$\dbinom 0 0 = 1 = \dbinom 1 1$ When $m > 0$ we also have by definition: :$\dbinom 0 m = 0 = \dbinom 1 {m + 1}$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P r$ is true, where $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{k \mathop = 0}^r \binom k m = \binom {r + 1} {m + 1}$ Then we need to show: :$\displaystyle \sum_{k \mathop = 0}^{r + 1} \binom k m = \binom {r + 2} {m + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{r + 1} \binom k m       | r = \sum_{k \mathop = 0}^r \binom k m + \binom {r + 1} m       | c =  }} {{eqn | r = \binom {r + 1} {m + 1} + \binom {r + 1} m       | c = Induction Hypothesis }} {{eqn | r = \binom {r + 2} {m + 1}       | c = Pascal's Rule }} {{end-eqn}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z, m \ge 0, n \ge 0: \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$ {{qed}}
{{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^n \paren {m + k}       | r = \frac {\paren {m + n}!} {m!}       | c =  }} {{eqn | r = n! \frac {\paren {m + n}!} {m! \, n!}       | c =  }} {{eqn | r = n! \binom {m + n} m       | c = {{Defof|Binomial Coefficient}} }} {{eqn | r = \binom {m + n} m \prod_{k \mathop = 1}^n k       | c =  }} {{end-eqn}} Hence the result, and note that for a bonus we have identified exactly what the divisor is:  :$\dbinom {m + n} m$ {{qed}}
Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$. Then: : $z^n \in \set {e^{2 i k \pi}: k \in \N_n}$ Hence $z^n = 1$. Now suppose $z^n = 1$. We determine the possible values of $z$ using Roots of Complex Number. Let $z = r e^{i \theta}$. Then $\cmod {z^n} = 1 \implies \cmod z = 1$. Similarly, we have $n \theta = 0 \bmod 2 \pi$. So $\theta = \dfrac {2 k \pi} n$ for $k \in \Z$. Hence the result. {{qed}} {{handwaving|This is a bit slapdash - anyone care to improve it?}}
From Complex Roots of Unity in Exponential Form: :$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$ where $U_n$ is the set of complex $n$th roots of unity. Let $\omega = e^{2 i \pi / n}$. Then we have: :$U_n = \set {\omega^k: k \in \N_n}$ that is: :$U_n = \set {\omega^0, \omega^1, \omega^2, \ldots, \omega^{n - 1} }$ Let $\omega^a, \omega^b \in U_n$. Then $\omega^a \omega^b = \omega^{a + b} \in U_n$. Either $a + b < n$, in which case $\omega^{a + b} \in U_n$, or $a + b \ge n$, in which case: {{begin-eqn}} {{eqn | l = \omega^a \omega^b       | r = \omega^{a + b}       | c =  }} {{eqn | r = \omega^{n + t}       | c = for some $t < n$ }} {{eqn | r = \omega^n \omega^t       | c =  }} {{eqn | r = \omega^t       | c = as $\omega^n = 1$ }} {{end-eqn}} So $U_n$ is closed under multiplication. We have that $\omega_0 = 1$ is the identity and that $\omega^{n - t}$ is the inverse of $\omega^t$. Finally we note that $U_n$ is generated by $\omega$. Hence the result, by definition of cyclic group, and from Cyclic Groups of Same Order are Isomorphic: :$U_n = \gen \omega \cong C_n$. {{qed}}
By the definition of inverse mapping: {{begin-eqn}} {{eqn | l = g \circ f       | r = I_S       | c =  }} {{eqn | r = h \circ f       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = f \circ g       | r = I_T       | c =  }} {{eqn | r = f \circ h       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = h       | r = h \circ I_T       | c =  }} {{eqn | r = h \circ \paren {f \circ g}       | c =  }} {{eqn | r = \paren {h \circ f} \circ g       | c = Composition of Mappings is Associative }} {{eqn | r = I_S \circ g       | c =  }} {{eqn | r = g       | c =  }} {{end-eqn}} So $g = h$ and the inverse is unique. {{qed}}
Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, they are invertible. So $\SL {n, K}$ is a subset of $\GL {n, K}$. Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$. Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$. As $\mathbf A$ is invertible we have that it has an inverse  $\mathbf A^{-1} \in \GL {n, K}$. From Determinant of Inverse Matrix: :$\map \det {\mathbf A^{-1} } = \dfrac 1 {\map \det {\mathbf A} }$ and so: :$\map \det {\mathbf A^{-1} } = 1$ So $\mathbf A^{-1} \in \SL {n, K}$. Also, from Determinant of Matrix Product: :$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B} = 1$ Hence the result from the Two-Step Subgroup Test. {{qed}}
Let $\mathbf I_n$ denote the unit matrix of order $n$ over $R$. The proof proceeds by induction. For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :$\map \det {\mathbf I_n} = 1_R$ By definition of Determinant of Order $1$: :$\begin {vmatrix} a_{1 1} \end {vmatrix} = a_{1 1}$ In this case $a_{1 1} = 1_R$. Thus $\map P 1$ is seen to hold. === Basis for the Induction === {{begin-eqn}} {{eqn | l = \map \det {\mathbf I_2}       | r = \begin {vmatrix} 1_R & 0_R \\ 0_R & 1_R \end {vmatrix}       | c =  }} {{eqn | r = 1_R \cdot 1_R - 0_R \cdot 0_R       | c = {{Defof|Determinant of Order 2}} }} {{eqn | r = 1_R       | c =  }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\map \det {\mathbf I_k} = 1_R$ from which it is to be shown that: :$\map \det {\mathbf I_{k + 1} } = 1_R$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \mathbf I_{k + 1}       | r = \begin {bmatrix} 1_R & 0_R \\ 0_R & \mathbf I_n \end {bmatrix}       | c = {{Defof|Unit Matrix}} }} {{eqn | ll= \leadsto       | l = \map \det {\mathbf I_{k + 1} }       | r = \begin {vmatrix} 1_R & 0_R \\ 0_R & \mathbf I_n \end {vmatrix}       | c = {{Defof|Determinant of Matrix}} }} {{eqn | r = \map \det {\mathbf I_k}       | c = Determinant with Unit Element in Otherwise Zero Row }} {{eqn | r = 1_R       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \map \det {\mathbf I_n} = 1_R$ {{qed}}
By definition of inverse matrix: :$\mathbf A \mathbf A^{-1} = \mathbf I_n$ where $\mathbf I_n$ is the unit matrix. By Determinant of Unit Matrix: :$\map \det {\mathbf I_n} = 1_K$ By Determinant of Matrix Product: :$\map \det {\mathbf A^{-1} } \map \det {\mathbf A} = \map \det {\mathbf A^{-1} \mathbf A}$ Hence the result. {{qed}}
Let $\pi = \begin{pmatrix} 1 & 2 \end{pmatrix}$ be a transposition. Let $\Delta_n$ be defined as Product of Differences. Then $\forall n \in \N_{>0}: \pi \cdot \Delta_n$ produces only one sign change in $\Delta_n$, that is, the one occurring in the factor $\paren {x_1 - x_2}$. Thus: :$\begin{pmatrix} 1 & 2 \end{pmatrix} \cdot \Delta_n = - \Delta_n$ and thus $\begin{pmatrix} 1 & 2 \end{pmatrix}$ is odd. From Conjugates of Transpositions: :$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}$ Thus as $\begin{pmatrix} 2 & k \end{pmatrix}$ is self-inverse: :$\begin{pmatrix} 1 & k \end{pmatrix} = \begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1}$ But from Parity of Conjugate of Permutation: :$\map \sgn {\begin{pmatrix} 2 & k \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & k \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$ Thus: :$\sgn {\begin{pmatrix} 1 & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & 2 \end{pmatrix} }$ and thus $\begin{pmatrix} 1 & k \end{pmatrix}$ is odd. Finally: {{begin-eqn}} {{eqn | l = \begin{pmatrix} h & k \end{pmatrix}       | r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}       | c = Conjugates of Transpositions }} {{eqn | r = \begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1}       | c = Transposition is Self-Inverse }} {{end-eqn}} But from Parity of Conjugate of Permutation: :$\map \sgn {\begin{pmatrix} 1 & h \end{pmatrix} \begin{pmatrix} 1 & k \end{pmatrix} \begin{pmatrix} 1 & h \end{pmatrix}^{-1} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$ Thus: :$\sgn {\begin{pmatrix} h & k \end{pmatrix} } = \sgn {\begin{pmatrix} 1 & k \end{pmatrix} }$ and thus $\begin{pmatrix} h & k \end{pmatrix}$ is odd. {{qed}}
Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$. By the definition of cardinality, $H$ is equivalent to $\N_n$. {{WLOG}} we can consider $S_n$ acting directly on $\N_n$. The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$. A permutation can be applied to $\N_n$ so that $i \to n$ for any $i$. Thus one can build an isomorphism to show the result for a general $i$. {{qed}}
Let $g \in G$. First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that: :$g * S \subseteq G$ Next we note: :$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s: s \in S} = S$ and so {{GroupActionAxiom|2}} is satisfied. Now let $g, h \in G$. We have: {{begin-eqn}} {{eqn | l = \paren {g \circ h} * S       | r = \paren {g \circ h} \circ S       | c =  }} {{eqn | r = \set {\paren {g \circ h} \circ s: s \in S}       | c =  }} {{eqn | r = \set {g \circ \paren {h \circ s}: s \in S}       | c =  }} {{eqn | r = g * \set {h \circ s: s \in S}       | c =  }} {{eqn | r = g * \paren {h \circ S}       | c =  }} {{eqn | r = g * \paren {h * S}       | c =  }} {{end-eqn}} and so {{GroupActionAxiom|1}} is satisfied. Hence the result. {{qed}}
First note that the identity $e$ satisfies $e^k = e$ and so $B$ is non-empty. Now assume that $a, b \in B$. Then: {{begin-eqn}} {{eqn | l=\left({ab^{-1} }\right)^k       | r=a^k \left({b^{-1} }\right)^k       | c=Power of Product in Abelian Group }} {{eqn | r=a^k \left({b^k}\right)^{-1}       | c=Powers of Group Elements }} {{eqn | r=ee^{-1}       | c=as $a^k = b^k = e \in B$ }} {{eqn | r=e       | c=Identity is Self-Inverse }} {{end-eqn}} Hence, by the One-Step Subgroup Test, $B$ is a subgroup of $A$. {{qed}} Category:Abelian Groups l27o0qcigojo3cxmvwlx9mpc6om1o7t