Main note: [[Free groups]] This is a short note generating sets. I mused earlier about what the *smallest* set that generates a given set is, and if what the smallest number of relations then must be. Such problems tend to be [incredibly hard to answer in general](https://mathoverflow.net/questions/76417/size-of-smallest-generating-set-of-a-group), and lie within the purview of **combinatorial group theory.** Still, there is a cute little concept that we can talk about in terms of minimal generating sets. Namely, we call an element a **nongenerator** to be an element that is not "important" when presenting a group via generators and relations, and thus can always be discarded if we're trying to find a small generating set. Formally, we define $g \in G$ to be a nongenerator if for every generating set $S$ containing $g$, the set $S\setminus \{g\}$ is also a generating set. We denote the set of all nongenerators by $\Phi(G)$ and we know that this is nonempty because the identity is always a nongenerator. What's more interesting is that $\Phi(G)$ turns out to be a subgroup of $G$, and is in fact the intersection of all the *maximal* subgroups of $G$ (a subgroup is called maximal if it's a proper subgroup of $G$ but is not contained in any other proper subgroup of $G$). For one direction of the proof, let $g$ be in every maximal subgroup, and suppose $S$ was a generating set containing $g$ such that $S\setminus \{g\} = S'$ is *not* a generating set. Now, the subgroup $\langle S' \rangle$ can't be , but it is contained in some maximal subgroup $M$^[technically this doesn't have to hold, but we won't consider the case where there are no maximal subgroups] (which can be $\langle S'\rangle$). By definition, $g \in M$, but then $S'$ can only generate $M$, not the whole space. This gives us a contradiction, so we know that the intersection of all maximal subgroups is a subgroup of nongenerators. Conversely, let $g$ be a nongenerator and a maximal subgroup not containing it. Then $M \cup \{g\}$ is a generating set for $G$ because it contains and is a strictly larger subgroup of $G$. But then $g$ can't be a nongenerator. So $g$ must have been in $M$ in the first place. This means that every nongenerator is in every maximal subgroup and thus $\Phi(G)$ is contained in the intersection of them all. This proof automatically gives us that $\Phi(G)$ must be a subgroup, but we can also see it directly. By definition a generating set allows taking inverses, so the inverse of a nongenerator must also be a nongenerator. Also, if $a$ and $b$ are both nongenerators, then so must be $ab$, loosely because we can always use the rest of the generating set to generate $a$ and $b$ which can be used to generate $ab$. Another interesting property about nongenerators is that any automorphism of $G$ sends a nongenerator to a nongenerator. This is quite easy to see because the image of a generating set under an automorphism is a generating set, so we can always remove the image of a nongenerator and the remaining thing is the image of a generating set. The reason this is interesting is because this means that any automorphism of $G$ sends to itself. Any subgroup $H$ of $G$ that is fixed by all automorphisms of $G$ is called a **characteristic** subgroup, and notably all characteristic subgroups are normal because conjugation $\phi_g(n) = gng^{-1}$ is always an automorphism for every $g$. The property that nongenerators are always sent to themselves means that $\Phi(G)$ is always characteristic and hence normal. Back to main note: [[Free groups]]