I am trying to prove the following.

let $G$ be a finitely generated abelian group, and $H<G$ a subgroup such that there exists a subgroup $K<G$ and we can write $G=H \oplus K$. Is it true that the minimal number of generators of H is strictly smaller than the minimal number of generators of $G$?

Clearly if G can not be written as a direct summand of $H$ then this is not true, just consider $G= \mathbb{Z}$ and $H=2\mathbb{Z}$.

I would like to prove it because I believe it can provide a simpler proof for the characterization of finitely generated abelian groups.

  • $\begingroup$ $\mathbb Z\big / 2\mathbb Z \oplus \mathbb Z\big / 3\mathbb Z $ is cyclic. $\endgroup$ – lulu Apr 21 at 18:54
  • 2
    $\begingroup$ Worth noting: "number of generators" is not well defined. I'm guessing you mean "minimal number of generators", but you should say so, $\endgroup$ – lulu Apr 21 at 18:55
  • $\begingroup$ Thank you for pointing that out. I will edit to correct it. $\endgroup$ – Charles Apr 21 at 19:00

No, it is not true. Consider $\mathbb{Z}_2\oplus\mathbb{Z}_3$. This has a generator $(1,1)$. Note that $$0\oplus\mathbb{Z}_3<\mathbb{Z}_2\oplus\mathbb{Z}_3 ,$$ and $$(\mathbb{Z}_2\oplus 0)\oplus(0\oplus\mathbb{Z_3})=\mathbb{Z}_2\oplus\mathbb{Z}_3.$$ However, $0\oplus\mathbb{Z}_3$ is generated by $(0,1).$


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