Methods to prove that a subset generates the whole group

What kind of methods exist to prove that a subset of a finitely generated abelian group G generates G

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How is the group specified (e.g. matrices, words with rewrite rules, operators on a space, etc.)? – Victor Miller Nov 17 '10 at 19:30

If you know the generators, represent each of the elements of your set as a linear combination of generators. Form a (non-square) matrix where row number $i$ consists of coefficients of element number $i$ in your set. Perform integer Gauss elimination procedure (i.e. you are allowed switching two rows, and subtracting/adding one row from/to another row). Eventually you will get a matrix in the row echelon form. Look how many 1's you have on the diagonal. If the number of 1's is the same as the number of generators, your set generates the whole group. Otherwise the answer is "no".

Edit: I forgot one more transformation in the Gauss elimination procedure: switching columns (these correspond to re-orderings of the set of generators of the group). Without it, the pivotal numbers will not be on the diagonal.

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The problem is that we have a INFINITE group, which is finitely generated. – Meriton Ibraimi Nov 17 '10 at 14:37
Yes, I assumed the group was infinite. So your set and the matrix can be infinite also (infinite number of rows, but each row is finite). Do I need to explain how to perform the Gauss elimination procedure on such a matrix? It is usually explained in the proofs that every f.g. Abelian group is a direct product of cyclic groups (or, which is almost the same, every f.g. module over a PID is a sum of cyclic modules). – Mark Sapir Nov 17 '10 at 14:45