# Calculate Smith normal form, cyclic group decomposition

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. $C_3=C_3+C_2$
• It's OK this time. The group is $A\simeq\mathbb{Z}_6\oplus\mathbb{Z}$. – ancientmathematician Apr 30 '18 at 10:37
• 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. – ancientmathematician Apr 30 '18 at 10:42
• 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 – user326441 Apr 30 '18 at 10:46
• 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 $X=0, 6Y=0, 0Z=0$. – ancientmathematician Apr 30 '18 at 11:05
• Hmm okay I'll need to think this over I haven't seen this before unfortunately haha – user326441 Apr 30 '18 at 11:09

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 $$d_1=1, d_2=6, d_3=0.$$