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I had the following finitely presented group:

$$G=\langle a,b: 2a+4b=0, 3b=0\rangle$$

We can use elementary operations by rows and columns and scalar multiplication using integers, so

$$\begin{pmatrix} 2 & 4 \\ 0 & 3 \end{pmatrix} \rightarrow \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $$

But I also get $$\begin{pmatrix} 2 & 4 \\ 0 & 3 \end{pmatrix} \rightarrow \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix} $$

Now to use the theorem the elements of the diagonal must divide, I mean

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \rightarrow\begin{pmatrix} d_1 & 0 \\ 0 & d_2 \end{pmatrix}$$ with $d_1|d_2$.

To work it out, I multiply by $2$ and $3$, to get the desired form in each matrix

$$ \begin{pmatrix} 2 & 0 \\ 0 & 6 \end{pmatrix} \text{, }\begin{pmatrix} 3 & 0 \\ 0 & 6 \end{pmatrix}$$

Am I doing something wrong or this means that the finite presentation of the group gives us the groups $\mathbb{Z}_2 \times \mathbb{Z}_6$ and $\mathbb{Z}_3 \times \mathbb{Z}_6$?

Edit:

The algorithm I'm trying to use is the following:

Theorem: If $A$ is a abelian group with $n$ generators $\{ a_1,\ldots , a_n\}$ and $m$ relations, $r_{i1}a_1+\cdots +r_{in}a_n=0$, $i=1,\cdots ,m$, then we consider the matrix $R=(r_{ij})_{m \times n}$. If $R$ is similar to the diagonal matrix:

$$\begin{pmatrix} d_1 & & & & & \\ & \ddots & & & & \\ & & d_s & & & \\ & & & 0 & & \\ & & & & \ddots & \\ & & & & & 0 \end{pmatrix}$$

By row and column operations and multiplication by integers, with $d_j\mid d_{j+1}$, $d_1\neq 0$, then

$A \cong \mathbb{Z}_{d_1} \oplus \cdots \oplus \mathbb{Z}_{d_s} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb{Z} $ (the sum of the $\mathbb{Z}$'s is $m-s$).

Shaun
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George
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  • From the given relations we quickly see that $b=-2a$, and therefore $G$ is generated by $a$ alone, hence cyclic (which is incompatible with either proposed answer). Can you be more specific about the algorithm you are trying to use and what specifically it accomplishes? – Greg Martin Jul 05 '21 at 21:02
  • Of course, I did the edit with the algotihm – George Jul 05 '21 at 22:06
  • Your diagonal form gives you that the group is isomorphic to $C_2\times C_3$ (since you reduced to the group $\langle a,b\mid 2a=3b=0\rangle$. But this group is cyclic of order $6$. – Arturo Magidin Jul 05 '21 at 22:33
  • But in the theorem the elements of the diagonal has the divisibility condition that do not hold for $d_1=2$ and $d_2=3$ – George Jul 05 '21 at 22:43
  • Right, so you haven't actually carried the algorithm out completely correctly. In particular, multiplying by an integer (other than $\pm1$) is not a legal row/column operation in this context, I believe. (Otherwise you could artificially inflate $d_s$ to any of its multiples, for example.) Further legal operations on $\begin{pmatrix} 2 & 0 \ 0 & 3 \end{pmatrix}$ would presumably bring it into the desired form $\begin{pmatrix} 6 & 0 \ 0 & 0 \end{pmatrix}$. – Greg Martin Jul 05 '21 at 23:12
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    @GregMartin: the desired form is $\mathrm{Diag}(1,6)$, not $\mathrm{Diag}(6,0)$; the latter corresponds to the group $\mathbb{Z}/6\mathbb{Z}\oplus\mathbb{Z}$. – Arturo Magidin Jul 06 '21 at 00:08
  • Not every sequence of elementary integral row and column operations will land you in a diagonal matrix of the desired form. You should check the algorithm for finding the Smith Normal Form and follow it, so that you transform your matrix into the desired form. For example, see here. – Arturo Magidin Jul 06 '21 at 00:17

1 Answers1

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You aren't done computing the Smith Normal Form; you still have work left to do. You should review the algorithm you have for computing the Smith Normal Form and follow it until it terminates. I'll do the final few steps ad hoc below, so the precise sequence may not match the algorithm you have/know/should have. Note in particular that you aren't allowed to multiply a row or column by anything other that $1$ or $-1$. You can exchange rows or columns; you can add integer multiples of one row to another, or one column to another; and you can multiply a row or column by $1$ or $-1$; these are "integral elementary row operations".

The gcd of the diagonal entries is $1$, so you'll need to do integral elementary row and column operations to get a first row and column with a $(1,1)$ entry equal to $1$ and zero in the other entries. We can achieve this by first adding first column to second column, then subtracting second row from first row, then do row and column exchanges to move the $1$ to the $(1,1)$ entry. Then use it to eliminate the $(2,1)$ entry, and then the $(1,2)$ entry. Below, something of the form $R_1\to$blah means "obtained by replacing row $1$ with the result of doing blah", etc.

$$\begin{align*} \left(\begin{array}{cc} 2 & 0\\ 0 & 3 \end{array}\right) &\to \left(\begin{array}{cc} 2 & 2\\ 0 & 3 \end{array}\right)&(C_2\to C_1+C_2)\\ &\to \left(\begin{array}{rr} 2 & 2\\ -2 & 1 \end{array}\right)&(R_2\to R_2-R_1)\\ &\to \left(\begin{array}{rr} 2 & 2\\ 1 & -2 \end{array}\right)&(C_1\leftrightarrow C_2)\\ &\to \left(\begin{array}{cc} 1 & -2\\ 2 & 2 \end{array}\right)&(R_1\leftrightarrow R_2)\\ &\to \left(\begin{array}{cc} 1 & -2\\ 0 & 6 \end{array}\right) &(R_2\to R_2-2R_1)\\ &\to \left(\begin{array}{cc} 1 & 0\\ 0 & 6 \end{array}\right)&(C_2\to C_2+2C_1) \end{align*}$$ which is of the desired form, and gives you that the resulting group is isomorphic to $\mathbb{Z}/1\mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z}\cong \mathbb{Z}/6\mathbb{Z}$.

Arturo Magidin
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