avewright/proofwiki-math huggingface dataset:
Theorem: Group of Order p^2 q has Normal Sylow p-Subgroup

Let $p$ and $q$ be prime numbers such that $p \ne q$.
Let $G$ be a group of order $p^2 q$.
Then $G$ has a normal Sylow $p$-subgroup.

Proof:
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides p^2 q$
where $\divides$ denotes divisibility.
Thus $n_p \in \set {1, q}$.
Suppose $p > q$.
Then:
:$q \not \equiv 1 \pmod p$
and so $n_p \ne q$.
Hence $n_p = 1$.
From Sylow p-Subgroup is Unique iff Normal it follows that this Sylow $p$-subgroup is normal.
{{qed|lemma}}
Suppose $q > p$.
Then $n_p$ could equal $q$ if $q \equiv 1 \pmod p$.
Let $n_q$ denote the number of Sylow $q$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_q \equiv 1 \pmod q$
and so $n_q$ is not a multiple of $q$.
From the Fifth Sylow Theorem:
:$n_q \divides p^2 q$
It follows that:
:$n_q \in \set {1, p, p^2}$
If $n_q = 1$ then it follows from Sylow p-Subgroup is Unique iff Normal that this Sylow $q$-subgroup is normal.
{{qed|lemma}}
As $p \not \equiv 1 \pmod q$, it follows that $n_q \ne p$.
The final possibility to be explored is when $n_q = p^2$.
Then:
{{begin-eqn}}
{{eqn | l = p^2
      | o = \equiv
      | r = 1
      | rr= \pmod q
      | c = 
}}
{{eqn | ll= \leadsto
      | l = q
      | o = \divides
      | r = p^2 - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = q
      | o = \divides
      | r = \paren {p + 1} \paren {p - 1}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = q
      | r = p + 1
      | c = as $q > p$
}}
{{end-eqn}}
But then $p$ and $q$ are consecutive prime numbers, and so:
:$p = 2, q = 3$
This gives the specific case:
:$\order G = 2^2 \times 3$
and the result follows from Normal Sylow p-Subgroups in Group of Order 12.
{{qed}}
Theorem: Group of Order 30 has Normal Cyclic Subgroup of Order 15

Let $G$ be of order $30$.
Then $G$ has a normal subgroup of order $15$ which is cyclic.

Proof:
By Group of Order 15 is Cyclic Group, any subgroup of $G$ of order $15$ is cyclic.
It remains to be proved that a subgroup of $G$ of order $15$ exists, and that it is normal.
Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_3 \equiv 1 \pmod 3$
and from the Fifth Sylow Theorem:
:$n_3 \divides 10$
where $\divides$ denotes divisibility.
It follows that $n_3 \in \set {1, 10}$.
Let $n_5$ denote the number of Sylow $5$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_5 \equiv 1 \pmod 5$
and from the Fifth Sylow Theorem:
:$n_5 \divides 30$
It follows that $n_5 \in \set {1, 6}$.
Suppose $n_3 = 1$.
Let $P$ denote the unique Sylow $3$-subgroup of $G$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
Thus we may form the quotient group $G / P$, which is of order $10$.
From Groups of Order 2p, $G / P$ has a unique Sylow $5$-subgroup, which we will denote $N / P$.
By the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.
As noted, $N$ is cyclic.
{{qed|lemma}}
Suppose $n_3 = 10$.
Let the Sylow $3$-subgroups of $G$ be denoted $P_1, P_2, \ldots, P_{10}$.
Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, \ldots, 10}, i \ne j$ is the trivial subgroup of $G$:
:$P_i \cap P_j = \set e$
Thus $G$ contains:
:The identity element $e$
:$20$ elements of order $3$, of which $2$ each are in $P_1, P_2, \ldots, P_{10}$
:$9$ more elements of $G$ which are in subgroups of $G$ different from the Sylow $3$-subgroups, and so have orders different from $3$.
Suppose $n_5 = 6$.
By a similar argument to the above, these $6$ Sylow $5$-subgroups contribute $24$ elements of order $5$ to $G$.
This is not possible because there are only $9$ elements of $G$ so far unaccounted for.
So if $G$ has $10$ Sylow $3$-subgroups, it can have only $1$ Sylow $5$-subgroup, which we will denote $Q$.
From Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.
Thus we may form the quotient group $G / Q$, which is of order $16$.
From Groups of Order 2p, $G / Q$ has a unique Sylow $3$-subgroup, which we will denote $N / Q$.
Again by the Correspondence Theorem, $N$ is a normal subgroup of $G$ with $15$ elements.
As noted, $N$ is cyclic.
{{qed|lemma}}
Both cases have been accounted for:
:that $G$ has $1$ Sylow $3$-subgroup
:that $G$ has $10$ Sylow $3$-subgroups.
Both cases result in $G$ having a normal subgroup of order $15$ which is cyclic.
{{qed}}
Theorem: Groups of Order 30/Lemma

Let $G$ be a group of order $30$.
Then $G$ is one of the following:
:The cyclic group $C_{30}$
:The dihedral group $D_{15}$
:Isomorphic to one of:
::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$

Proof:
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 = {{Group-axiom|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}}
Theorem: Normal Subgroup of Group of Order 24

Let $G$ be a group of order $24$.
Then $G$ has either:
:a normal subgroup of order $8$
or:
:a normal subgroup of order $4$.

Proof:
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}}
Theorem: Group of Order 35 is Cyclic Group

Let $G$ be a group whose order is $35$.
Then $G$ is cyclic.

Proof:
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 Group of Order $p q$ is Cyclic can be applied.
{{Qed}}
Theorem: Group of Order 35 is Cyclic Group

Let $G$ be a group whose order is $35$.
Then $G$ is cyclic.

Proof:
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 {{Group-axiom|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 {{Group-axiom|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}}
Theorem: Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup

Let $G$ be a group of order $105$.
Then $G$ has either:
:exactly one normal Sylow $5$-subgroup
or:
:exactly one normal Sylow $7$-subgroup.

Proof:
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}}
Theorem: Group of Order 105 has Normal Cyclic Subgroup of Index 3

Let $G$ be a group of order $105$.
Then $G$ has a normal cyclic subgroup $N$ such that:
:$\index G N = 3$
where $\index G N$ denotes the index of $N$ in $G$.

Proof:
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}}
Theorem: Diagonal Relation is Reflexive

The diagonal relation $\Delta_S$ on a set $S$ is a reflexive relation in $S$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x \in S
      | l = x
      | r = x
      | c = {{Defof|Equals}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, x}
      | o = \in
      | r = \Delta_S
      | c = {{Defof|Diagonal Relation}}
}}
{{end-eqn}}
So $\Delta_S$ is reflexive.
Theorem: Diagonal Relation is Symmetric

The diagonal relation $\Delta_S$ on a set $S$ is a symmetric relation in $S$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = \tuple {x, y}
      | o = \in
      | r = \Delta_S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = {{Defof|Diagonal Relation}}
}}
{{eqn | ll= \leadsto
      | l = y
      | r = x
      | c = Equality is Symmetric
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \in
      | r = \Delta_S
      | c = {{Defof|Diagonal Relation}}
}}
{{end-eqn}}
So $\Delta_S$ is symmetric.
Theorem: Diagonal Relation is Transitive

The diagonal relation $\Delta_S$ on a set $S$ is a transitive relation in $S$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = \tuple {x, y}
      | o = \in
      | r = \Delta_S \land \tuple {y, z} \in \Delta_S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y \land y = z
      | c = {{Defof|Diagonal Relation}}
}}
{{eqn | ll= \leadsto
      | l = x
      | r = z
      | c = Equality is Transitive
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, z}
      | o = \in
      | r = \Delta_S
      | c = {{Defof|Diagonal Relation}}
}}
{{end-eqn}}
So $\Delta_S$ is transitive.
Theorem: Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup

Let $G$ be a group of order $56$.
Then $G$ has either:
:exactly one normal Sylow $2$-subgroup
or:
:exactly one normal Sylow $7$-subgroup.

Proof:
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}}
Theorem: Subgroup of Direct Product is not necessarily Direct Product of Subgroups

Let $G$ and $H$ be groups.
Let $G \times H$ denote the direct product of $G$ and $H$.
Let $K$ be a subgroup of $G \times H$.
Then it is not necessarily the case that $K$ is of the form:
:$G' \times H'$
where:
:$G'$ is a subgroup of $G$
:$H'$ is a subgroup of $H$.

Proof:
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}}
Theorem: Groups of Order 30

Let $G$ be a group of order $30$.
Then $G$ is one of the following:
:The cyclic group $C_{30}$
:The dihedral group $D_{15}$
:The group direct product $C_5 \times D_3$
:The group direct product $C_3 \times D_5$

Proof:
First we introduce a lemma:
Theorem: Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with unity $1_R$.
Then:
:$\norm {\,\cdot\,}$ is non-Archimedean $\implies \forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$.
where:
$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} }$

Proof:
Let $\norm {\,\cdot\,}$ be non-Archimedean.
Then by the definition of a non-Archimedean norm, for $n \in \N$:
{{begin-eqn}}
{{eqn | q = \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}}
Theorem: Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with unity $1_R$.
Then:
:$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1 \implies \norm {\,\cdot\,}$ is non-archimedean
where:
:$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{n \text { times} }$

Proof:
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 = {{Norm-axiom-mult|3}}
}}
{{eqn | r = \sum_{i \mathop = 0}^n \norm {\binom n i \cdot 1_R} \norm x^i
      | c = {{Norm-axiom-mult|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|Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition|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:
:$\ds \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} \max \set {\norm x, 1} = \max \set {\norm x, 1}$
By Inequality Rule for Real Sequences:
:$\norm {x + 1_R} \le \max \set {\norm x, 1}$
The result follows.
Theorem: Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 1

Let $y \ne 0_R$ where $0_R$ is the zero of $R$.
Then:
:$\norm {x + y} \le \max \set {\norm x, \norm y} \iff \norm {x y^{-1} + 1_R} \le \max \set {\norm {x y^{-1} }, 1}$

Proof:
{{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 = {{Norm-axiom-mult|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|D}}
}}
{{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}}
Theorem: Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2

Then for all $i$, $0 \le i \le n$:
:$\norm x^i \le \max \set {\norm x^n , 1}$

Proof:
If $\norm x > 1$ then for all $i$, $0 \le i \le n$:
:$\norm x^i \le \norm x^n \le \max \set {\norm x^n, 1}$
If $\norm x \le 1$ then for all $i$, $0 \le i \le n$:
:$\norm x^i \le 1 \le \max \set {\norm x^n, 1}$
In either case for all $i$, $0 \le i \le n$:
:$\norm x^i \le \max \set {\norm x^n , 1}$
{{qed}}