avewright/proofwiki-math huggingface dataset:
Theorem: Subgroups of Cartesian Product of Additive Group of Integers

Let $\struct {\Z, +}$ denote the additive group of integers.
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z \times \Z, +}$ denote the Cartesian product of $\struct {\Z, +}$ with itself.
The subgroups of $\struct {\Z \times \Z, +}$ are not all of the form:
:$\struct {m \Z, +} \times \struct {n \Z, +}$
where $\struct {m \Z, +}$ denotes the additive group of integer multiples of $m$.

Proof:
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by:
:$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$
which is a group isomorphism.
{{explain|Prove the above statement}}
Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a free abelian group of rank $2$.
Therefore, any subgroup generated by a singleton, for example, $\set {\tuple {x, 0}: x \in \Z}$ is a subgroup not in the form $\struct {m \Z, +} \times \struct {n \Z, +}$.
{{refactor|Implement "singly generated" as a definition page in its own right and then link to it, hence keep this page clean|level = medium}}
{{qed}}
Theorem: Equivalence Relation on Symmetric Group by Image of n is Congruence Modulo Subgroup

Let $S_n$ denote the symmetric group on $n$ letters $\set {1, \dots, n}$.
Let $\sim$ be the relation on $S_n$ defined as:
:$\forall \pi, \tau \in S_n: \pi \sim \tau \iff \map \pi n = \map \tau n$
Then $\sim$ is an equivalence relation which is congruence modulo a subgroup.
{{explain|Work needed to be done to explain exactly what is happening here.}}

Proof:
We claim that $\sim$ is left congruence modulo $S_{n - 1}$, the symmetric group on $n - 1$ letters $\set {1, \dots, n - 1}$.
Notice that every element of $S_{n - 1}$ fixes $n$.
For all $\pi, \tau \in S_n$ such that $\pi \sim \tau$:
{{begin-eqn}}
{{eqn | l = \map {\paren {\pi^{-1} \circ \tau} } n
      | r = \map {\pi^{-1} } {\map \tau n}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map {\pi^{-1} } {\map \pi n}
      | c = Definition of $\sim$
}}
{{eqn | r = \map {\paren {\pi^{-1} \circ \pi} } n
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = n
      | c = {{Defof|Inverse Element}}
}}
{{end-eqn}}
so $\pi^{-1} \circ \tau$ fixes $n$ as well.
This shows that $\pi^{-1} \circ \tau \in S_{n - 1}$.
By definition of Left Congruence Modulo Subgroup:
:$\pi \equiv^l \tau \pmod {S_{n - 1} }$
Now we show the converse.
Suppose $\pi \equiv^l \tau \pmod {S_{n - 1} }$.
Then $\pi^{-1} \circ \tau \in S_{n - 1}$.
Hence $\map {\paren {\pi^{-1} \circ \tau} } n = n$.
Then:
{{begin-eqn}}
{{eqn | l = \map \pi n
      | r = \map \pi {\map {\paren {\pi^{-1} \circ \tau} } n}
}}
{{eqn | r = \map {\paren {\pi \circ \pi^{-1} \circ \tau} } n
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map \tau n
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = \pi
      | o = \sim
      | r = \tau
}}
{{end-eqn}}
Therefore $\sim$ and left congruence modulo $S_{n - 1}$ are equivalent.
The fact that $\sim$ is an equivalence relation follows from Left Congruence Modulo Subgroup is Equivalence Relation.
{{qed}}
Theorem: Order of Additive Group of Integers Modulo m

Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
The order of $\struct {\Z_m, +_m}$ is $m$.

Proof:
By definition, the order of a group is the cardinality of its underlying set.
By definition, the underlying set of $\struct {\Z_m, +_m}$ is the set of residue classes $\Z_m$:
:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
From Cardinality of Set of Residue Classes, $\Z_m$ has $m$ elements.
Hence the result.
{{qed}}
Theorem: Order of Multiplicative Group of Reduced Residues

Let $\struct {\Z'_m, \times_m}$ denote the multiplicative group of reduced residues modulo $m$.
The order of $\struct {\Z'_m, \times_m}$ is $\map \phi m$, where $\phi$ denotes the Euler $\phi$ function.

Proof:
By definition, the order of a group is the cardinality of its underlying set.
By definition, the underlying set of $\struct {\Z'_m, \times_m}$ is the reduced residue system  $\Z'_m$:
:$\Z'_m = \set {\eqclass {a_1} m, \eqclass {a_2} m, \ldots, \eqclass {a_{\map \phi m} } m}$
where:
:$\forall k: a_k \perp m$
From Cardinality of Reduced Residue System, $\Z'_m$ has $\map \phi m$ elements.
Hence the result.
{{qed}}
Theorem: Circle Group is Uncountably Infinite

The circle group $\struct {K, \times}$ is an uncountably infinite group.

Proof:
From Quotient Group of Reals by Integers is Circle Group, $\struct {K, \times}$ is isomorphic to the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$.
But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the half-open interval $\hointr 0 1$.
A real interval is uncountable by Real Numbers are Uncountable.
{{explain|No, the above link Real Numbers are Uncountable just shows that $\R$ is uncountable, not an interval.}}
Hence the result.
{{qed}}
{{MissingLinks|results in here}}
Theorem: Subgroup of Symmetric Group that Fixes n

Let $S_n$ denote the symmetric group on $n$ letters.
Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
:$\map \pi n = n$
Then:
:$H = S_{n - 1}$
and the index of $H$ in $S_n$ is given by:
:$\index {S_n} H = n$

Proof:
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters.
Let $\pi \in S_{n - 1}$.
Then $\pi$ is a permutation on $n - 1$ letters.
Hence $\pi$ is also a permutation on $n$ letters which fixes $n$.
So $S_{n - 1} \subseteq H$.
Now let $\pi \in H$.
Then $\pi$ is a permutation on $n$ letters which fixes $n$.
That is, $\pi$ is a permutation on $n - 1$ letters.
Thus $\pi \in S_{n - 1}$.
So we have that $H = S_{n - 1}$.
We also have that $S_{n - 1}$ is a group, and:
:$\forall \rho \in S_{n - 1}: \rho \in S_n$
So $S_{n - 1}$ is a subset of $S_n$ which is a group.
Hence $S_{n - 1}$ is a subgroup of $S_n$ by definition.
Then we have:
{{begin-eqn}}
{{eqn | l = \index {S_n} {S_{n - 1} }
      | r = \dfrac {\order {S_n} } {\order {S_{n - 1} } }
      | c = {{Defof|Index of Subgroup}}
}}
{{eqn | r = \dfrac {n!} {\paren {n - 1}!}
      | c = Order of Symmetric Group
}}
{{eqn | r = \dfrac {n \paren {n - 1}!} {\paren {n - 1}!}
      | c = {{Defof|Factorial}}
}}
{{eqn | r = n
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Order of Product of Abelian Group Elements Divides LCM of Orders of Elements

Let $G$ be an abelian group.
Let $a, b \in G$.
Then:
:$\order {a b} \divides \lcm \set {\order a, \order b}$
where:
:$\order a$ denotes the order of $a$
:$\divides$ denotes divisibility
:$\lcm$ denotes the lowest common multiple.

Proof:
Let $\order a = m, \order b = n$.
Let $c = \lcm \set {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = c
      | r = r m
      | c = for some $r \in \Z$
}}
{{eqn | r = s n
      | c = for some $s \in \Z$
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \paren {a b}^c
      | r = a^c b^c
      | c = Power of Product of Commuting Elements in Semigroup equals Product of Powers
}}
{{eqn | r = a^{r m} b^{s n}
      | c = 
}}
{{eqn | r = \paren {a^m}^r \paren {b^n}^s
      | c = 
}}
{{eqn | r = e^r e^s
      | c = {{Defof|Order of Group Element}}
}}
{{eqn | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {a b}
      | o = \divides
      | r = c
      | c = Element to Power of Multiple of Order is Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Elements of Abelian Group whose Order Divides n is Subgroup

Let $G$ be an abelian group whose identity element is $e$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer .
Let $G_n$ be the subset of $G$ defined as:
:$G_n = \set {x \in G: \order x \divides n}$
where:
:$\order x$ denotes the order of $x$
:$\divides$ denotes divisibility.
Then $G_n$ is a subgroup of $G$.

Proof:
From Identity is Only Group Element of Order 1:
:$\order e = 1$
and so from One Divides all Integers:
:$\order e \divides n$
Thus $G_n \ne \O$.
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = G_n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order x
      | o = \divides
      | r = n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {x^{-1} }
      | o = \divides
      | r = n
      | c = Order of Group Element equals Order of Inverse
}}
{{eqn | ll= \leadsto
      | l = x^{-1}
      | o = \in
      | r = G_n
      | c = 
}}
{{end-eqn}}
Let $a, b \in G_n$ such that $\order a = r, \order b = s$.
Then from Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:
:$\order {a b} = \divides \lcm \set {a, b}$
But $r \divides n$ and $s \divides n$ by definition of $G_n$.
Therefore, by definition of lowest common multiple:
:$\order {a b} \divides n$
Thus we have:
:$G_n \ne \O$
:$x \in G_n \implies x^{-1} \in G_n$
:$a, b \in G_n \implies a b \in G_n$
and the result follows by the Two-Step Subgroup Test.
{{qed}}
Theorem: Groups of Order 4

There exist exactly $2$ groups of order $4$, up to isomorphism:
:$C_4$, the cyclic group of order $4$
:$K_4$, the Klein $4$-group.

Proof:
From Existence of Cyclic Group of Order n we have that one such group of order $4$ is the cyclic group of order $4$:
This is exemplified by the additive group of integers modulo $4$, whose Cayley table can be presented as:
{{:Modulo Addition/Cayley Table/Modulo 4}}
From Group whose Order equals Order of Element is Cyclic, any group with an element of order $4$ is cyclic.
From Cyclic Groups of Same Order are Isomorphic, no other groups of order $4$ which are not isomorphic to $C_4$ can have an element of order $4$.
{{qed|lemma}}
From Order of Element Divides Order of Finite Group, any other group of order $4$ must have elements of order $2$.
We have the Klein $4$-group, whose Cayley table can be presented as:
{{:Klein Four-Group/Cayley Table}}
and is seen to have that property.
{{qed|lemma}}
We have that Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic.
{{qed|lemma}}
It remains to be shown that the Klein $4$-group is the only groups of order $4$ whose elements are all of order $2$ (except the identity).
Let the Cayley table be populated as far as can be directly established:
:$\begin{array}{c|cccc}
  & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e &   &   \\
b & b &   & e &   \\
c & c &   &   & e \\
\end{array}$
Consider $a b$.
As $a^2 = e$, $a b \ne e$.
As $a e = a$, $a b \ne a$.
As $e b = b$, $a b \ne b$.
It follows that $a b = c$.
Hence we have:
:$\begin{array}{c|cccc}
  & e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & e & c &   \\
b & b &   & e &   \\
c & c &   &   & e \\
\end{array}$
and the rest of the table is completed by following the result that Group has Latin Square Property.
{{qed}}
Theorem: Finite Group whose Subsets form Chain is Cyclic P-Group

Let $G$ be a group.
Let $G$ be such that its subgroups form a chain.
Then $G$ is a cyclic $p$-group.

Proof:
Suppose $G$ is not a $p$-group.
Then there exist two distinct primes $p_1, p_2$.
By Cauchy's Group Theorem, there exist subgroups $H, K$ such that:
:$\order H = p_1$
:$\order K = p_2$
That is:
:$H = \gen a$
:$k = \gen b$
for some $a, b \in G: a \ne b$, where:
:$\order a = p_1$
:$\order b = p_2$
and so both $H \nsubseteq K$ and $K \nsubseteq H$
Thus, by definition, the subgroups of $G$ do not form a chain.
It follows by the Rule of Transposition that $G$ is a $p$-group.
It remains to be shown that $G$ is cyclic.
Suppose to the contrary that $G$ is not cyclic.
Then it is not the case that $G$ is generated by a single element of $G$.
Thus there exist $a, b \in G, a \ne b$ such that $H = \gen a$ and $K = \gen b$ are subgroups of $G$.
But as $a \in H, b \in K, a \notin K, b \notin H$ it follows that $H \nsubseteq K$ and $K \nsubseteq H$.
It follows by the Rule of Transposition that $G$ is cyclic.
{{qed}}
Theorem: Subgroup of Circle Group Generated by Distinct Roots of Unity

Let $K$ be the circle group.
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $d = \lcm \set {m, n}$ be the least common multiple of $m$ and $n$.
Let $\alpha$ be a primitive $n$th root of unity.
Let $\beta$ be a primitive $m$th root of unity.
Let $\gamma$ be a primitive $d$th root of unity.
Let $H = \gen {\alpha, \beta}$ be the subgroup of $K$ generated by $\alpha, \beta$.
Then $H = \gen \gamma$.

Proof:
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Theorem: Multiplicative Group of Complex Roots of Unity is Subgroup of Circle Group

Let $n \in \Z$ be an integer such that $n > 0$.
Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity.
Let $\struct {K, \times}$ denote the circle group.
Then $\struct {U_n, \times}$ is a subgroup of $\struct {K, \times}$.

Proof:
By definition of the multiplicative group of complex $n$th roots of unity:
:$U_n := \set {z \in \C: z^n = 1}$
By definition of the circle group:
:$K = \set {z \in \C: \cmod z = 1}$
By Modulus of Complex Root of Unity equals 1:
:$\forall z \in U_n: \cmod z = 1$
Thus:
:$U_n \subseteq K$
We further have that the operation $\times$ on both $U_n$ and $K$ is complex multiplication.
Finally, from Roots of Unity under Multiplication form Cyclic Group, we have that $\struct {U_n, \times}$ is a group.
The result follows by definition of subgroup.
{{qed}}
Category:Circle Group
Category:Multiplicative Groups of Complex Roots of Unity
lfvp2f3yr0fc3tttjne1u8897x8k13i
Theorem: Intersection of Multiplicative Groups of Complex Roots of Unity

Let $\struct {K, \times}$ denote the circle group.
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $c = \lcm \set {m, n}$ be the lowest common multiple of $m$ and $n$.
Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity.
Let $\struct {U_m, \times}$ denote the multiplicative group of complex $m$th roots of unity.
Let $H = U_m \cap U_n$.
Then $H = U_c$.

Proof:
{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
Theorem: Direct Product of Normal Subgroups is Normal

Let $G$ and $G'$ be groups.
Let:
:$H \lhd G$
:$H' \lhd G'$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {H \times H'} \lhd \paren {G \times G'}$
where $H \times H'$ denotes the group direct product of $H$ and $H'$

Proof:
Let $\tuple {x, x'} \in G \times G'$ and $\tuple {y, y'} \in H \times H'$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, x'} \tuple {y, y'} \tuple {x, x'}^{-1}
      | r = \tuple {x, x'} \tuple {y, y'} \tuple {x^{-1}, x'^{-1} }
      | c = 
}}
{{eqn | r = \tuple {x y x^{-1}, x' y' x'^{-1} }
      | c = 
}}
{{eqn | o = \in
      | r = H \times H'
      | c = {{Defof|Normal Subgroup}}
}}
{{end-eqn}}
Hence:
:$\paren {H \times H'} \lhd {G \times G'}$
{{qed}}
Theorem: Normalizer of Rotation in Dihedral Group

Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \alpha}$ denote the normalizer of the singleton containing the rotation element $\alpha$.
Then:
:$\map {N_{D_n} } {\set \alpha} = \gen \alpha$
where $\gen \alpha$ is the subgroup generated by $\alpha$.

Proof:
By definition, the normalizer of $\set \alpha$ is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$
That is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$
First let $g = \alpha^k$ for some $k \in \Z$.
Then:
:$\alpha \alpha^k = \alpha^k \alpha$
which includes $k = 0$, that is $e$.
Thus:
:$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$
Now let $g = \alpha^k \beta$.
Suppose $g \alpha = \alpha g$.
Then:
{{begin-eqn}}
{{eqn | l = \alpha^k \beta \alpha
      | r = \alpha \alpha^k \beta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \beta \alpha
      | r = \alpha \beta
      | c = Cancellation Laws
}}
{{end-eqn}}
But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$.
Hence the result:
:$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$
Hence the result, by definition of generator of subgroup.
{{qed}}
Category:Dihedral Groups
Category:Normalizers
l8t1v1hdfwid63js3d7czck2lhjh8dr
Theorem: Normalizer of Reflection in Dihedral Group

Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \beta}$ denote the normalizer of the singleton containing the reflection element $\beta$.
Then:
:$\map {N_{D_n} } {\set \beta} = \begin{cases} \set {e, \beta} & : n \text { odd} \\ \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta} & : n \text { even} \end{cases}$

Proof:
By definition, the normalizer of $\set \beta$ is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$
That is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$
First let $g = \beta^k$ for $k \in \set {0, 1}$.
Then:
:$\beta \beta^k = \beta^k \beta$
Thus:
:$\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$
Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$.
Suppose $g \alpha = \alpha g$.
Then:
{{begin-eqn}}
{{eqn | l = \alpha^j \beta^k \beta
      | r = \beta \alpha^j \beta^k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^j \beta \beta^k
      | r = \beta \alpha^j \beta^k
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^j \beta
      | r = \beta \alpha^j
      | c = Cancellation Laws
}}
{{eqn | ll= \leadsto
      | l = \alpha^j \beta
      | r = \alpha^{n - j} \beta 
      | c = Product of Generating Elements of Dihedral Group
}}
{{eqn | ll= \leadsto
      | l = \alpha^j
      | r = \alpha^{n - j}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \alpha^2 j
      | r = \alpha^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 j
      | r = n
      | c = 
}}
{{end-eqn}}
Thus when $n$ is odd, there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
But when $n$ is even such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
Hence the result.
{{qed}}
Category:Dihedral Groups
Category:Normalizers
or2apavxjj0cj0ux4v60y43iu4pql3h
Theorem: Normalizer of Subgroup of Symmetric Group that Fixes n

Let $S_n$ denote the symmetric group on $n$ letters.
Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
:$\map \pi n = n$
The normalizer of $H$ is given by:
:$\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$

Proof:
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$.
By definition of normalizer:
:$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$
We have from Group is Normal in Itself that:
:$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$
and so:
:$S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$
It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$.
This will be done by demonstrating that:
: $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
where $\setminus$ denotes set difference.
Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$.
Thus:
:$\map \rho n \ne n$
and so:
:$\exists a \in S_n, a \ne n: \map \rho a = n$
for some $a \in S_{n - 1}$.
Then:
:$\map {\rho^{-1} } n = a$
Let $\pi \in S_{n - 1}$ such that:
: $\map \pi a = b$
for some $b \ne a$.
As $\rho$ is a permutation, $\rho$ is by definition a bijection.
Hence:
:$\map \rho b \ne n$
We have:
{{begin-eqn}}
{{eqn | l = \map {\rho \pi \rho^{-1} } n
      | r = \map {\rho \pi} a
      | c = as $\map {\rho^{-1} } n = a$
}}
{{eqn | r = \map \rho b
      | c = as $\map \pi a = b$
}}
{{eqn | o = \ne
      | r = n
      | c = as $\map \rho b \ne n$
}}
{{end-eqn}}
Thus $\rho \pi \rho^{-1}$ does not fix $n$.
That is:
:$\rho \pi \rho^{-1} \notin S_{n - 1}$
Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that:
:$S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
So from Intersection with Complement is Empty iff Subset:
:$\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$
and so by definition of set equality:
:$\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
{{qed}}
Theorem: Center of Quaternion Group

Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group.
Let $\map Z {\Dic 2}$ denote the center of $\Dic 2$.
Then:
:$\map Z {\Dic 2} = \set {e, a^2}$

Proof:
By definition, the center of $\Dic 2$ is:
:$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$
We are given that:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
We have that $\Dic 2$ is generated by $\alpha$ and $\beta$.
Thus:
:$x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$
Let $x \in \map Z {\Dic 2}$.
We have that $x$ can be expressed in the form:
:$x = a^i b^j$
for $i \in \set {0, 1, 2, 3}$ and $j \in \set {0, 1}$.
As $x \in \map Z {\Dic 2}$, we have:
{{begin-eqn}}
{{eqn | l = x a
      | r = a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^i b^j a
      | r = a^{i + 1} b^j
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b^j a
      | r = a b^j
      | c = 
}}
{{end-eqn}}
For $j = 1$ this means:
{{begin-eqn}}
{{eqn | l = a b
      | r = b a
      | c = 
}}
{{eqn | r = a^{-1} \beta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | r = a^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2
      | r = e
      | c = 
}}
{{end-eqn}}
But the order of $a$ is $4$, not $2$, and hence:
:$a^2 \ne e$
So if $x \in \map Z {\Dic 2}$ it follows that $x$ has to be in the form:
:$x = a^i$
for some $i \in \Z_{\ge 0}$.
Again, as $x \in \map Z {\Dic 2}$, we have:
{{begin-eqn}}
{{eqn | l = x b
      | r = b x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^i b
      | r = b a^i
      | c = 
}}
{{eqn | r = a^{-i} b
      | c = Product of Generating Elements of Quaternion Group
}}
{{eqn | ll= \leadsto
      | l = a^i
      | r = a^{-i}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^{2 i}
      | r = e
      | c = 
}}
{{eqn | r = a^4
      | c = {{Defof|Order of Group Element}}
}}
{{end-eqn}}
So either $i = 0$ or $2 i = 4$, as $0 \le i \le 4$.
If $i = 0$ then $x = a^0 = e$.
If $2 i = 4$ then:
:$x = a^{4 / 2} = a^2$
Hence the result.
{{qed}}
Theorem: Product of Generating Elements of Quaternion Group

Let $Q = \Dic 2$ be the quaternion group:
:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
Then for all $k \in \Z_{\ge 0}$:
:$b a^k = a^{-k} b$

Proof:
The proof proceeds by induction.
For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition:
:$b a^k = a^{-k} b$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = b a^0
      | r = b e
      | c = 
}}
{{eqn | r = e b
      | c = 
}}
{{eqn | r = a^{-0} b
      | c = 
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.