Definition. A set $\{x_1,...,x_n\}$ of non zero elements in an Abelian group is independent if, whenever there are integers $m_1,...,m_r$ with $m_1 x_1 + \cdots + m_r x_r = 0$, then each $m_i x_i$ is zero. (Introduction to the Theory of Groups, 4th edition, by Rotman, Springer-Verlag, p.127.)

I want to know:

In a finite Abelian group, is a minimal generating set necessarily an independent set?

I have tried to prove it but I have failed. Perhaps if the group is primary (a p-group for some prime p)?

More precisely, suppose I have a set that generates G. By removing elements from the set one at a time, can I get an independent set which still generates G?


This is not always the case not even for $p$-groups. Take the direct sum of a cyclic group of order $p$ and one of order $p^2$. Let them be generated by $g$ and $h$ respectively. This would be an independent generating set. But now consider $g+h$ and $h$. This is still minimal (of course, as the group is not cyclic), yet not independent .

It is of course true for direct sums of groups of order $p$ a prime (the notion of independence coincides with linear independent over the field with $p$ elements, and this is just linear algebra).

It remains true for direct sums of groups of order $p^k$ a prime power. (This is not very hard to see.)

And, I believe, but did not fully think it through that these are all the finite abelian groups for which it is true.

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  • $\begingroup$ Actually a cyclic group of ord $pq$ would be another example where it is true. (I refrain from saying that now it should be complete. If/when I actually figure out the complete list I'll update this.) $\endgroup$ – quid Apr 3 '16 at 12:57

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