algorithm we can make below group on natural numbers: this group is in accordance with Collatz graph ,
let
∀
m
,
n
∈
N
,
{\displaystyle \forall m,n\in \mathbb {N} ,\qquad }
{
m
⋆
1
=
m
(
4
m
)
⋆
(
4
m
−
2
)
=
1
=
(
4
m
+
1
)
⋆
(
4
m
−
1
)
(
4
m
−
2
)
⋆
(
4
n
−
2
)
=
4
m
+
4
n
−
5
(
4
m
−
2
)
⋆
(
4
n
−
1
)
=
4
m
+
4
n
−
2
(
4
m
−
2
)
⋆
(
4
n
)
=
{
4
m
−
4
n
−
1
4
m
−
2
>
4
n
4
n
−
4
m
+
1
4
n
>
4
m
−
2
3
m
=
n
+
1
(
4
m
−
2
)
⋆
(
4
n
+
1
)
=
{
4
m
−
4
n
−
2
4
m
−
2
>
4
n
+
1
4
n
−
4
m
+
4
4
n
+
1
>
4
m
−
2
(
4
m
−
1
)
⋆
(
4
n
−
1
)
=
4
m
+
4
n
−
1
(
4
m
−
1
)
⋆
(
4
n
)
=
{
4
m
−
4
n
+
2
4
m
−
1
>
4
n
4
n
−
4
m
4
n
>
4
m
−
1
2
m
=
n
(
4
m
−
1
)
⋆
(
4
n
+
1
)
=
{
4
m
−
4
n
−
1
4
m
−
1
>
4
n
+
1
4
n
−
4
m
+
1
4
n
+
1
>
4
m
−
1
3
m
=
n
+
1
(
4
m
)
⋆
(
4
n
)
=
4
m
+
4
n
−
3
(
4
m
)
⋆
(
4
n
+
1
)
=
4
m
+
4
n
(
4
m
+
1
)
⋆
(
4
n
+
1
)
=
4
m
+
4
n
+
1
N
=
⟨
2
⟩
=
⟨
4
⟩
{\displaystyle {\begin{cases}m\star 1=m\\(4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\(4m-2)\star (4n-2)=4m+4n-5\\(4m-2)\star (4n-1)=4m+4n-2\\(4m-2)\star (4n)={\begin{cases}4m-4n-1&4m-2>4n\\4n-4m+1&4n>4m-2\\3&m=n+1\end{cases}}\\(4m-2)\star (4n+1)={\begin{cases}4m-4n-2&4m-2>4n+1\\4n-4m+4&4n+1>4m-2\end{cases}}\\(4m-1)\star (4n-1)=4m+4n-1\\(4m-1)\star (4n)={\begin{cases}4m-4n+2&4m-1>4n\\4n-4m&4n>4m-1\\2&m=n\end{cases}}\\(4m-1)\star (4n+1)={\begin{cases}4m-4n-1&4m-1>4n+1\\4n-4m+1&4n+1>4m-1\\3&m=n+1\end{cases}}\\(4m)\star (4n)=4m+4n-3\\(4m)\star (4n+1)=4m+4n\\(4m+1)\star (4n+1)=4m+4n+1\\\mathbb {N} =\langle 2\rangle =\langle 4\rangle \end{cases}}}
Collatz conjecture , let
C
1
:=
{
(
m
,
2
m
)
∣
m
∈
N
}
{\displaystyle C_{1}:=\{(m,2m)\mid m\in \mathbb {N} \}}
{
e
C
1
=
(
1
,
2
)
∀
m
,
n
∈
N
,
(
m
,
2
m
)
⋆
C
1
(
n
,
2
n
)
=
(
m
⋆
n
,
2
(
m
⋆
n
)
)
(
m
,
2
m
)
−
1
=
(
m
−
1
,
2
×
m
−
1
)
that
m
⋆
m
−
1
=
1
C
1
=
⟨
(
2
,
4
)
⟩
=
⟨
(
4
,
8
)
⟩
≃
Z
{\displaystyle {\begin{cases}e_{C_{1}}=(1,2)\\\\\forall m,n\in \mathbb {N} ,\,(m,2m)\star _{C_{1}}(n,2n)=(m\star n,2(m\star n))\\(m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad {\text{that}}\quad m\star m^{-1}=1\\\\C_{1}=\langle (2,4)\rangle =\langle (4,8)\rangle \simeq \mathbb {Z} \end{cases}}}
and
C
2
:=
{
(
3
m
−
1
,
2
m
−
1
)
∣
m
∈
N
}
{\displaystyle C_{2}:=\{(3m-1,2m-1)\mid m\in \mathbb {N} \}}
{
e
C
2
=
(
2
,
1
)
∀
m
,
n
∈
N
,
(
3
m
−
1
,
2
m
−
1
)
⋆
C
2
(
3
n
−
1
,
2
n
−
1
)
=
(
3
(
m
⋆
n
)
−
1
,
2
(
m
⋆
n
)
−
1
)
(
3
m
−
1
,
2
m
−
1
)
−
1
=
(
3
×
m
−
1
−
1
,
2
×
m
−
1
−
1
)
that
m
⋆
m
−
1
=
1
C
2
=
⟨
(
5
,
3
)
⟩
=
⟨
(
11
,
7
)
⟩
≃
Z
{\displaystyle {\begin{cases}e_{C_{2}}=(2,1)\\\\\forall m,n\in \mathbb {N} ,\,(3m-1,2m-1)\star _{C_{2}}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\(3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad {\text{that}}\quad m\star m^{-1}=1\\\\C_{2}=\langle (5,3)\rangle =\langle (11,7)\rangle \simeq \mathbb {Z} \end{cases}}}
and let
C
:=
C
1
⊕
C
2
{\displaystyle C:=C_{1}\oplus C_{2}}
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
Question : What are maximal subgroups of
C
{\displaystyle C}
— Preceding unsigned comment added by 89.45.54.114 (talk ) 06:22, 21 May 2018 (UTC) [ reply ]
