Let $G$ be the quotient group $G=\mathbb{Z}^5/N$, where $N$ is generated by $(6,0,-3,0,3)$ and $(0,0,8,4,2)$. Recognize $G$ as a product of cyclic groups.
Honestly, I do not know how to solve these type of problems. But I know that this is somehow an application of Fundamental theorem of finitely generated abelian groups. That theorem states an existence of such a product as $\mathbb{Z}^r\times \mathbb{Z}_{n_1}\times … \times \mathbb{Z}_{n_s}$, but does not states a way to find $r,n_1,…,n_s$. I know how to use this theorem for a finite abelian group. But could not find a way to solve these type of problems even in a book. Could somebody explain me?
Best Answer
As noted by SpamIAm, the key to this will be the Smith normal form of a matrix. Basically, we write $N$ as the image of a matrix and then find an easier, similar matrix for which the solution is clearer.
Since we have 2 generators for $N$ we have a surjective homomorphism $\mathbb{Z}^2\to N$ given by the matrix $A$ of generators for $N$: $$A=\begin{pmatrix}6&0\\0&0\\-3&8\\0&4\\3&2\end{pmatrix}$$ and $N=A\mathbb{Z}^2$. Now, this matrix its self is kind of messy, but it turns out that we can use a similar matrix and if we use a simple enough matrix the problem becomes easier.
Now you find the Smith normal form by using row and column operations to simplify as much as possible. I got the following matrix but you should check my work yourself: $$A=\begin{pmatrix}6&0\\0&0\\-3&8\\0&4\\3&2\end{pmatrix} \sim\begin{pmatrix}3&2\\0&0\\-3&8\\0&4\\6&0\end{pmatrix} \sim\begin{pmatrix}1&2\\0&0\\-11&8\\-4&4\\6&0\end{pmatrix} \sim\begin{pmatrix}1&0\\0&0\\0&30\\0&12\\0&-12\end{pmatrix} \sim\begin{pmatrix}1&0\\0&6\\0&0\\0&0\\0&0\end{pmatrix}=A'$$ Now we have $\mathbb{Z}^5/N=\mathbb{Z}^5/(A\mathbb{Z}^2)\cong\mathbb{Z}^5/(A'\mathbb{Z}^2)$. But it should be easy to see that $A'\mathbb{Z}^2=\mathbb{Z}\times(6\mathbb{Z})$ which gives us $$G\cong \dfrac{\mathbb{Z}^5}{\mathbb{Z}\times(6\mathbb{Z})}=\mathbb{Z}^3\times\mathbb{Z}_6$$