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Can someone please check my working for the following problem?

Let $A$ be the abelian group generated by elements $x,y,z$ with relations $7x+5y+2z=0, 3x+3y=0, 13x+11y+2z=0$. Decompose $A$ as a direct sum of cyclic groups.

$$ M = \begin{pmatrix} 7 & 5 & 2 \\ 3 & 3 & 0 \\ 13 & 11 & 2 \end{pmatrix} $$

$$\begin{pmatrix} 7 & 5 & 2 \\ 3 & 3 & 0 \\ 13 & 11 & 2 \end{pmatrix} \sim \begin{pmatrix} 7 & 5 & 2 \\ 3 & 3 & 0 \\ -1 & 1 & -2 \end{pmatrix} \sim \begin{pmatrix} 1 & -1 & 2 \\ 3 & 3 & 0 \\ 7 & 5 & 2 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 3 & 6 & -6 \\ 7 & 12 & -12 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 6 & -6 \\ 0 & 12 & -12 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 6 & -6 \\ 0& 0 & 0 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$

So $A = \mathbb{Z}/1\mathbb{Z} \oplus \mathbb{Z}/6\mathbb{Z} \oplus \mathbb{Z} / 0\mathbb{Z}$ based on the above Smith normal form which I'm not confident about.

The 6 steps taken are the following:

  1. $R_3=R_3-2R_1$
  2. Swap $R_3$ and $-R_1$
  3. $C_2=C_2+C_1$ and $C_3=C_3-2C_1$
  4. $R_2=R_2-3R_1$ and $R_3=R_3-7R_1$
  5. $R_3=R_3-2R_2$
  6. $C_3=C_3+C_2$
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  • $\begingroup$ It's OK this time. The group is $A\simeq\mathbb{Z}_6\oplus\mathbb{Z}$. $\endgroup$ – ancientmathematician Apr 30 '18 at 10:37
  • $\begingroup$ However the question says "decompose into cyclic subgroups" and all you've done is find what the cyclic subgroups are isomorphic to - rather than finding the cyclic subgroups as it requires. $\endgroup$ – ancientmathematician Apr 30 '18 at 10:42
  • $\begingroup$ I'm sorry I don't think I know how to do that. Would you mind explaining what happens next or what I need to do? I'd really appreciate it $\endgroup$ – user326441 Apr 30 '18 at 10:46
  • $\begingroup$ If you keep track of the changes you are making to the generators $x,y,z$ when you do column operations you'll find that with $X:=x-y+2z$, $Y:=y-z$, $Z:=z$ we have that $X,Y,Z$ generate $A$ with $X=0, 6Y=0, 0Z=0$. $\endgroup$ – ancientmathematician Apr 30 '18 at 11:05
  • $\begingroup$ Hmm okay I'll need to think this over I haven't seen this before unfortunately haha $\endgroup$ – user326441 Apr 30 '18 at 11:09
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Since you want to check your answer, there is a direct way to obtain Smith normal form; define inductively $$d_1d_2\cdots d_k=\mbox{gcd of $k\times k$ minors of matrix under consideration}.$$ Then the Smith normal form is $$\begin{bmatrix} d_1 & & \\ & d_2 & \\ & & d_3\end{bmatrix}.$$ So $d_1=1$ because it is $gcd(7,5,..)=1$; next $d_1d_2=gcd(6,-6,12,-12)=6$, and $d_1d_2d_3=\det A=0.$ Thus $$d_1=1, d_2=6, d_3=0.$$

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  • $\begingroup$ Wow thank you so much! I didn't know you could do that. Do you know how to use that to find the cyclic subgroups needed for this problem? $\endgroup$ – user326441 May 1 '18 at 2:45

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