I am trying to prove that the minimum size of a minimal generating set of a finite abelian group $G$, denoted $d(G)$, where $G=C_{d_{1}} \times \dots \times C_{d_{k}}$ for $d_{i} \mid d_{i+1}$, is $k$. Note that I am using minimal generating set to mean no redundant generators. I have proven so far that $k=\max\{d(P)\}$, where each $P$ is a Sylow $p$-subgroup of $G$ for each prime $p \mid G$.
- To finish my proof of the main result, it suffices to show that $d(H) \leq d(G)$ for $H \leq G$. I know this does not hold for non-abelian groups, but I think it is true for abelian groups, although I am having a hard time proving it. Is this true, and if so could someone please point me to how I could prove this?
- If this is not true, could someone please point me to how I could continue the proof that $d(G)=k$?
Thank you for your help.
I believe I have managed to work out the proof by showing that every Sylow $p$-subgroup of $G$ is isomorphic to a quotient of $G$. I then used the fact that $d(G/N) \leq d(G)$, which follows by definition. This showed that $d(G) \geq k$. I also showed that $d(G) \leq k$ by using the fact that $d(H \times K) \leq d(H)+d(K)$.