Let there be two natural numbers n,m.

let there be d = greatest common divisor of m and n - gcd(m,n)

and l = least common multiple of m and n - lcm(m,n)

I need to prove that CnXCm isomprphic to ClXCd (Cm Cn Cl Cd are all cyclic groups)

I have tried to see what happens if I look at m and n as products of prime numbers but I am kind of stuck around that idea without knowing where to take it.

I also think I should use the fact that gcd(m,n)*lcm(m,n) = m*n but also, can't figure out where to take it where to take it.

Another thing, if the gcd(m,n) = 1 I know that CmXCn is cyclic and the order of it is mn. I thought maybe this fact could help me somehow (m*n = l*d), but I don't know how, because it is possible that gcd(m,n) > 1

can somebody push me towards the right path to solution?

thanks! ]]>

The general solution of the system $Ax=\begin{bmatrix}

1\\

3

\end{bmatrix}$ is $x=\begin{bmatrix}

1\\

0

\end{bmatrix}+ \lambda \begin{bmatrix}

0\\

1

\end{bmatrix}$. I want to find the matrix $A$.

I have done the following so far:

$$x=\begin{bmatrix}

1\\

0

\end{bmatrix}+ \lambda \begin{bmatrix}

0\\

1

\end{bmatrix}=\begin{bmatrix}

1\\

\lambda

\end{bmatrix}.$$

Let $A=\begin{bmatrix}

a_{11} & a_{12}\\

a_{21} & a_{22}

\end{bmatrix}$.

$$\begin{bmatrix}

a_{11} & a_{12}\\

a_{21} & a_{22}

\end{bmatrix} \begin{bmatrix}

1\\

\lambda

\end{bmatrix}=\begin{bmatrix}

1\\

3

\end{bmatrix} \Leftrightarrow \begin{bmatrix}

a_{11}+ \lambda a_{12}\\

a_{21}+\lambda a_{22}

\end{bmatrix}=\begin{bmatrix}

1\\

3

\end{bmatrix} \Leftrightarrow \begin{Bmatrix}

a_{11}+\lambda a_{12}=1 \\

a_{21}+\lambda a_{22}=3

\end{Bmatrix} \Leftrightarrow \begin{Bmatrix}

a_{11}=1-\lambda a_{12} \\

a_{21}=3-\lambda a_{22}

\end{Bmatrix}. $$

So $A$ is the following matrix:

$$A=\begin{bmatrix}

1-\lambda a_{12} & a_{12}\\

3-\lambda a_{22} & a_{22}

\end{bmatrix}.$$

Is everything right? Can we get more information or is this sufficient? (Thinking) ]]>