An interesting object defined for a finite group G

Question

Given a finite group $G$, we define $u(G):= \min\{|S| : S \subseteq G, \,\, \langle S\rangle = G \,\}$. We fix a group of order $n$.

1. Prove that if $H$ is a normal subgroup of $G$, then $u(G) \le u(H) + u(G/H)$.
2. Prove that if $ n = p^r$ for some prime number $p$, then $u(G) \le r$.
3. Prove that, if $n = 2100$ and $G$ is abelian, then $u(G) \le 2$.

I tried several ways to solve the first two points but without success. For the third one I know, up to isomorphisms, which are all the abelian groups of order 2100 and consequently determine one or two elements that generate these groups and thus arrive at the thesis, but this way suggests that there would be a better strathegy, perhaps using the first two points.

Thank you in advance for any answer or advice on the resolution!

Answer 1

I will give brief steps here, and you may fill in details yourself.

For 1:

Suppose $h_1, \dots, h_k$ generate $H$ and $\sigma_1, \dots, \sigma_t$ generate $G/H$. Take any lift $g_i \in G$ of $\sigma_i$. Try to show that $h_1, \dots, h_k, g_1, \dots, g_t$ generate $G$.

For 2:

Prove by induction that you can find $k$ elements in $G$ that generates a subgroup of order $\geq p^k$.

Alternatively, use the fact that a finite p -group cannot be simple unless it has order p and then apply 1. to a proper normal subgroup of $G$.

For 3:

I don't see any way of using the previous two points. You may use the fundamental theorem of finite abelian groups to classify the possible structures of $G$, and then use Chinese remainder theorem to find generators.