I need to prove the following:
Let $G$ be a group, generated by the elements $g_1, \ldots ,g_n $. In other words, a subgroup $\langle g_1, \ldots, g_n \rangle$ consists of all different possible products of the form:
$$g_{i_1}^{l_1}g_{i_2}^{l_2}\cdots g_{i_k}^{l_k},$$
where $g_{i_j}\in \{g_1,\ldots g_n\}$ and $l_i \in \mathbb{Z} $. A group $G$ is $n$-generated, if there exist $n$ elements $g_1 ,\ldots,g_n \in G $, so that $\langle g_1,\ldots,g_n \rangle = G $.
a) Let $g_1,\ldots,g_k \in G$ be nontrivial elements of group $G$ and let $g_{k+1} \in G$ be an element, such that $g_{k+1}\notin \langle g_1, \ldots, g_k\rangle$. Show that:
$$\langle g_1, \ldots,g_k\rangle \leq \langle g_1,\ldots, g_k,g_{k+1}\rangle \space \space \text{ and } \space \space |\langle g_1,\ldots, g_k, g_{k+1}\rangle |\geq 2|\langle g_1,\ldots, g_k \rangle |.$$
b) Let $G$ be a finite group of order $n$. Show that $G$ is $k$-generated, where $k \leq \log_2(n)$ .
My attempt:
a) Def. A subset $H \subset G$ is a subgroup of $G$, if $H$ forms a group for the binary (group) operation defined on $G$.
Both $\langle g_1,\ldots g_k\rangle=G_k$ and $\langle g_1,\ldots g_k,g_{k+1}\rangle=G_{k+1}$ (already) form groups, so all we need to show here is that $\langle g_1,\ldots g_k\rangle \subset\langle g_1,\ldots g_k, g_{k+1}\rangle$. But this is clearly the case, since if we take an element $g \in \langle g_1,\ldots g_k, g_{k+1}\rangle$ of the form $g=\ldots g_{k}^{l_k} g_{k+1} ^{l_{k+1}}$, then $g \in \langle g_1,\ldots g_k, g_{k+1}\rangle $ but $g \notin\langle g_1,\ldots g_k\rangle$.
Because if $g\in \langle g_1,\ldots g_k\rangle$ this would imply that $g_{k+1}\in \langle g_1,\ldots g_k\rangle$, since $\langle g_1,\ldots g_k\rangle$ forms a group and needs to be closed under the group operation. This contradicts $g_{k+1} \notin \langle g_1,\ldots g_k\rangle $. So we have $\langle g_1,\ldots g_k\rangle \subset\langle g_1,\ldots g_k, g_{k+1}\rangle$ and $\langle g_1,\ldots g_k\rangle \leq \langle g_1,\ldots g_k,g_{k+1}\rangle$.
For the second part: $| G_{k+1}= \langle g_1,\ldots, g_k, g_{k+1} \rangle | \geq 2 |G_k= \langle g_1,\ldots, g_k \rangle |$, we know from Lagrange's theorem that for the number of right cosets of $G_{k}$ in $G_{k+1}$, i.e. $|G_{k+1}:G_k|=|G_{k+1}|/|G_k|\in \mathbb{N}$.
Since clearly $|G_{k+1}|\neq |G_k|$ and (consequently) $|G_{k+1}|/|G_k|\neq 1$, we mush have $|G_{k+1}|/|G_k|\geq2$.
b) Here we form a chain of subgroups $1=G_0<G_1<\ldots<G_k$, so that $G_i=\langle g_1, \ldots g_i\rangle$. We now use the above argument that $|G_i:G_{i-1}|=|G_i|/|G_{i-1}|\geq 2$ iteratively: $$|G_k| \geq 2 |G_{k-1}|$$ $$\vdots$$ $$|G_1|\geq 2 |G_0|,$$ or in other words: $|G_k| \geq 2^k |G_0|=2^k $. Now $|G_k|=n$ and taking the logarithm of the last equation we get: $\log_2(n)\geq k$.
My question:
Is my reasoning here ok? Did I miss some steps in the proofs? Maybe someone can suggest a more compact way to prove this.
Related questions:
Show that the number of nonisomorphic finite groups of order $n$ is at most $n^{n^2}$ (see answer below)
