Consider a finite $2$-generated group $G$, not cyclic. Fix $a\in G$. Let's denote by $J_a$ the set of elements of $G$ that do not generate $G$ together with $a$, i.e. $$J_a=\{g\in G: \langle g,a \rangle \neq G\}.$$ Now, is it true that $J_a$ is a subgroup?
- $1_G\in J_a$,
- If $g\in J_a$, then $g^{-1}\in J_a$,
- If $g,h\in J_a$, assume that $\langle gh, a\rangle=G$, then $a$ and $gh$ are respectively in two maximal subgroups of $G$, call them $H$ and $K$, such that $\{gh,a\}\notin H\cap K$. So, from this point I can not argue to obtain that $gh\in J_a$...
Well, if $J_a$ is a subgroup, who is this object, for every $a\in G$? Of course, the union of maximal subgroups containing $a$ is in $J_a$.
For example, take the group $A_5$ and fix $a=(12345)\in A_5$. By calculations, I have found that $J_{(12345)}=N_G(\langle a \rangle)$, which is maximal in $A_5$. How can explain this fact?
As I noted above, $J_a$ is exactly the union of the maximal subgroups containing $a$. In your example, because for each $5$-cycle $\sigma$ there is a unique maximal subgroup of $A_5$ that contains $\sigma$, namely its normalizer, your calculation follows.
However, in general $J_a$ is not a subgroup. Consider $G=S_5$, which is $2$-generated (e.g., by $(12)$ and $(12345)$). Let $a=(12)$. Then $J_a$ contains the one point stabilizers of $3$, $4$, and $5$ (so, any permutation that fixes at least one point from $\{3,4,5\}$; in particular, it contains all transpositions), but $J_a\neq G$. Hence, $S_5=\langle J_a\rangle\neq J_a$, so $J_a$ is not a subgroup.