Prove that $x$ is in all maximal subgroups of a group $G$ iff $x \in X \subseteq G$, where $X$ generates $G$, then $X \setminus \{x\}$ generates $G$ Show that $x \in G$ lies in the intersection of all maximal subgroups of G if and only if it has the following property: if $X \subseteq G$ contains $x$ and generates $G$, then $X\setminus\{x\}$ generates $G$.
Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.
So the first part of the exercise is alright. If $x \in \bigcap(M_i)$, where $M_i$ is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by $X\setminus\{x\}$, so $G = \langle {X} \rangle = \langle{X}\setminus\{x\}\rangle$.
For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group. $\langle V_4 \setminus \{a\}\rangle = \langle 1, b, c \enspace | \enspace b^2 = c^2 = 1, bc = cb = a \rangle $, clearly generates $ \{1,a,b,c\}$, however $ a \notin \langle b \rangle, \langle c \rangle$, so the converse is not always true?
Where have I gone wrong?
 A: Your definition of maximal subgroup is wrong.

A maximal subgroup $H$ of $G$ is a subgroup such that $H\ne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $H\subsetneq K\subsetneq G$.

Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $x\in X$, where $X$ is a subset of $G$. Suppose $\langle X\setminus\{x\}\rangle\ne G$; then this subgroup is contained in a maximal subgroup $H$ and $x\in H$ by assumption, so $\langle X\rangle=\bigl<\langle X\setminus\{x\}\rangle\cup\{x\}\bigr>\subseteq H$. Therefore $\langle X\rangle\ne G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $X\subseteq G$ containing $x$, if $\langle X\rangle=G$, then $\langle X\setminus\{x\}\rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $x\notin H$, then $\langle H\cup\{x\}=G$, so, being $x$ a nongenerator, also $H=\langle (H\cup\{x\})\setminus\{x\}\rangle=G$, which is a contradiction.
In the Klein group $V=\{1,a,b,c\}$, the maximal subgroups are $\{1,a\}$, $\{1,b\}$ and $\{1,c\}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that $\{a,b,c\}$ generates $V$ but also $\{b,c\}$ generates $V$ doesn't mean $a$ is a nongenerator. This set $\{a,b,c\}$ is just one of the sets of generators containing $a$. Indeed, consider $X=\{a,b\}$. Then $\langle X\rangle=V$, but $\langle X\setminus\{a\}\rangle=\langle\{b\}\rangle=\{1,b\}\ne V$. Thus $a$ is not a nongenerator.
