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May 21

Collatz conjecture is almost done!

Following is a method for solving Collatz conjecture: regarding to this algorithm we can make below group on natural numbers: this group is in accordance with Collatz graph,

let $\mathrm{\forall }m,n\in \mathbb{N},$ $\{\begin{array}{c}m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\{\begin{array}{cc}4m-4n-1& 4m-2>4n\\ 4n-4m+1& 4n>4m-2\\ 3& m=n+1\end{array}\\ (4m-2)\star (4n+1)=\{\begin{array}{cc}4m-4n-2& 4m-2>4n+1\\ 4n-4m+4& 4n+1>4m-2\end{array}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\{\begin{array}{cc}4m-4n+2& 4m-1>4n\\ 4n-4m& 4n>4m-1\\ 2& m=n\end{array}\\ (4m-1)\star (4n+1)=\{\begin{array}{cc}4m-4n-1& 4m-1>4n+1\\ 4n-4m+1& 4n+1>4m-1\\ 3& m=n+1\end{array}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star (4n+1)=4m+4n+1\\ \mathbb{N}=⟨2⟩=⟨4⟩\end{array}$

now I am making a group in accordance with Collatz conjecture, let $C_1:=\{(m,2m)\mid m\in \mathbb{N}\}$ is a group with: $\{\begin{array}{c}{e}_{C_1}=(1,2)\\ \\ \mathrm{\forall }m,n\in \mathbb{N},(m,2m){\star }_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m{)}^{-1}=(m^{-1},2\times m^{-1})\text{that}m\star m^{-1}=1\\ \\ C_1=⟨(2,4)⟩=⟨(4,8)⟩\simeq \mathbb{Z}\end{array}$

and $C_2:=\{(3m-1,2m-1)\mid m\in \mathbb{N}\}$ is a group with: $\{\begin{array}{c}{e}_{C_2}=(2,1)\\ \\ \mathrm{\forall }m,n\in \mathbb{N},(3m-1,2m-1){\star }_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1{)}^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\text{that}m\star m^{-1}=1\\ \\ C_2=⟨(5,3)⟩=⟨(11,7)⟩\simeq \mathbb{Z}\end{array}$.

and let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$.

Question: What are maximal subgroups of $C$?

Thanks in advance! (I really have insufficient time, hence I need your help!) — Preceding unsigned comment added by 89.45.54.114 (talk) 06:22, 21 May 2018 (UTC)