The task states: "Find a integer matrix $A \in Mat_{3 \times 4} \mathbb Z$ which presents commutative group $\mathbb Z \times \mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_4$ and find it's Smith normal form". My thought is that SMT of such matrix should be $$ \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ \end{pmatrix} $$ but I have trouble finding matrix itself. Am I right about SMT form? How do I find such a matrix?
Presentation of commutative group by integer matrix
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$\begingroup$
linear-algebra
abelian-groups
group-presentation
1 Answers
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Your matrix below is fine for $4$ generators $$\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ \end{pmatrix} $$ but it's not in Smith normal form because we don't have $2 \mid 3 \mid 4$.
Since $\mathbb Z \times \mathbb Z_2 \times \mathbb Z_3 \times \mathbb Z_4 \cong \mathbb Z \times \mathbb Z_2 \times \mathbb Z_{12}$, the Smith normal form will be $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 12 \\ \end{pmatrix} $$ Note that we now need only $3$ generators.
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0I don't have $A$ - it should be the size 3x4, not 4x4. – 2017-02-05
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0This can't be right. The theorem about SMT states that exist such invertible $P,Q$ for $A$, so that $P^{-1}AQ = S$, where $S$ is matrix in SMT. Because $A$ is 3x4, $Q$ must be 4x4, so $S$ must be also 4x4 matrix. – 2017-02-05
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0Scratch that, SMT should be in 3x4 form, because we are multiplying 3x3 x 3x4 x 4x4 matrices. – 2017-02-05