Express a quotient of a free abelian group as a direct sum of cyclic groups

Question

So I'm studying for my algebra quals by going over some old quals from the school. One problem asks the following. Let $G$ denote the free abelian group on $\{ x , y , z \}$ with the relations $2x + 4y + 6z = 4x + 4y + 4z = 2x - 4y - 10z = 0$. The problem wants me to express $G$ as a direct sum of cyclic groups, and I just flat out don't know where to start with this. About the extent of the thoughts I had were the following: That $x , y, z$ would probably not be the easiest thing to work with, and that $\mathbb{Z}_2$ might be a factor on account of all the coefficients in the imposed relations being even. But I didn't know how to take either of these ideas to the bank. Another idea I had was that perhaps I could put this in a matrix and do some kind of row-reduction process, but it didn't seem to get me anywhere and I don't know what the idea would've been anyways.

I'm at a loss on how to do this problem. I did find this post, but the answer seems to refer to another answer that has since been deleted. One answer says something about the Smith form of a matrix, but I don't know what matrix he's taking the Smith form of, nor how to read from that matrix the structure of the group. I also don't know what a Smith form is, but that's something I imagine I can figure out.

If anyone can explain to me how to solve this problem, I'd really appreciate it! Thanks in advance.

Answer 1

This response is NOT a correct answer. as pointed out in the comments. The group $G$ indeed has some torsion elements. Please see the second answer I posted.

I will leave this here, just in case someone else makes a similar mistake.

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If we solve the 3 x 3 system $$\left\{\begin{array} {l} 2x+4y+6z=0\\4x+4y+4z= 0\\2x−4y−10z=0\end{array}\right.$$ using elementary row operations, we find that $y = -2x$ and $z =x$. So equivalently, $G$ is the free abelian group generated by $x, y, z$ with $y = -2x$ and $z =x$. This means that $G$ is just generated by $x$ and $G \cong \mathbb Z$.

Answer 2

First note that your abelian group $G$ is isomorphic to a cokernel of the additive map $$\varphi=\begin{bmatrix}2&4&2\\4&4&-4\\6&4&-10\end{bmatrix}:\Bbb Z^3\to\Bbb Z^3$$ that's there exists an additive map $\varepsilon:\Bbb Z^3\to G$ satifying $\operatorname{Im}\varphi=\operatorname{Ker}\varepsilon$ and \begin{align} &\varepsilon(1,0,0)=x& &\varepsilon(0,1,0)=y& &\varepsilon(0,0,1)=z \end{align} Diagonalize $\varphi$ according to Smith normal form which gives the factorization $\eta\circ\varphi\circ\xi=\sigma$ where \begin{align} &\eta=\begin{bmatrix}1&0&0\\2&-1&0\\1&-2&1\end{bmatrix}& &\xi=\begin{bmatrix}1&-2&3\\0&1&-2\\0&0&1\end{bmatrix}& &\sigma=\begin{bmatrix}2&0&0\\0&4&0\\0&0&0\end{bmatrix} \end{align} that's $$\begin{bmatrix}1&0&0\\2&-1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}2&4&2\\4&4&-4\\6&4&-10\end{bmatrix}\begin{bmatrix}1&-2&3\\0&1&-2\\0&0&1\end{bmatrix}=\begin{bmatrix}2&0&0\\0&4&0\\0&0&0\end{bmatrix}$$ Since $\sigma$ is diagonal, it's cokernel is given by the projection $\pi$ onto the abelian group $H=\Bbb Z/2\Bbb Z\times\Bbb Z/4\Bbb Z\times\Bbb Z$. According to the commutative diagram below

we get a group isomorphism $\psi:G\to H$ satisfying \begin{align} &\psi(x)=(1,2,1)& &\psi(y)=(0,-1,-2)& &\psi(z)=(0,0,1) \end{align} which are the columns of the matrix $\eta$.

Answer 3

If we use $\mathbb Z$-row operations on the 3 x 3 system $$\left\{\begin{array}l2x+4y+6z=0\\4x+4y+4z=0\\2x−4y−10z=0\\\end{array} \right.$$ we can "eliminate" a relation and eventually be left with $4y + 8z = 0$ and $2x - 2z =0$*. These $\mathbb Z$-row operations say that $\langle 2x + 4y + 6z, 4x + 4y + 4z, 2x - 4y - 10z\rangle = \langle 2x - 2z, 4y + 8z\rangle$ in the free abelian group on $x, y, z$ and furthermore $G$ is the free abelian group on $x,y, z$ modulo $\langle 2x - 2z, 4y + 8z\rangle$

Now to compute the decomposition of $G$: First we note that the free abelian group on $x, y, z$ can be generated by the three elements: $x, x - z, y+2z$. This is because $y = (y+2z) + 2(x-z) - 2x$ and $z = x - (x-z)$. So $$G \cong \frac{\mathbb Zx \oplus \mathbb Zy \oplus \mathbb Zz}{\langle 2x - 2z, 4y + 8z\rangle} \cong \frac{\mathbb Zx\oplus \mathbb Z(x-z) \oplus \mathbb Z(y + 2z)}{\langle 2x - 2z, 4y + 8z \rangle}\\ \cong {\mathbb Zx} \oplus \frac{\mathbb Z(x-z)}{2\mathbb Z(x-z)}\oplus \frac{\mathbb Z(y+2z)}{4\mathbb Z(y+2z)} \cong \mathbb Z \oplus \frac{\mathbb Z}{2\mathbb Z}\oplus \frac{\mathbb Z}{4\mathbb Z}$$

*Note these relations do not give $y = -2x$ and $x =z$ as I incorrectly said in my first answer, because we cannot necessarily ``divide by 2" as @FabioLucchini pointed out.