Wikipedia:Reference desk/Archives/Mathematics/2018 May 21

= 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 $$\forall m,n\in\mathbb N,\qquad$$ $$\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2> 4n\\ 4n-4m+1 & 4n> 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2> 4n+1\\ 4n-4m+4 & 4n+1> 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1> 4n\\ 4n-4m & 4n> 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1> 4n+1\\ 4n-4m+1 & 4n+1> 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star (4n+1)=4m+4n+1\\ \mathbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$$

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: $$\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\mathbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\simeq\mathbb Z\end{cases}$$

and $$C_2:=\{(3m-1,2m-1)\mid m\in\mathbb N\}$$ is a group with: $$\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\mathbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\simeq\mathbb Z\end{cases}$$.

and let $$C:=C_1\oplus C_2$$ 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)