Number of nonidentity Elements Contained in Conjugates of $M$ 
Prove that if $M$ is a maximal group of $G$ that is not normal in $G$, then the number of nonidentity elements of $G$ that are contained in conjugates of $M$ is at most $(|M|-1) |G:M|$.

Here is my reasoning: First, since $M$ is maximal but not normal, and $N \le N_G(M) \le G$, then $M =N_G(M)$. Clearly every $x \in M \setminus \{1\}$ is contained in a conjugate of $M$--namely, $xMx^{-1}$, which means that $|M|-1$ constitute some of the nonidentity elements we are endeavoring to count. Now, if $g \in G \setminus M = G \setminus N_G(M)$, then $gMg^{-1} \neq M$, which means there exists an $x \in gMg^{-1} \setminus M$ or $x M \setminus gMg^{-1}$. Since we our goal is to find an upper, we want to overcount by assuming that every such $x$ is in $gMg^{-1} \setminus M$. Again, for the sake of overcounting, we can assume that each $g \in G \setminus M$ produces a unique coset. This implies that the upper bound on the number of nonidentity elements contained in a conjugate of $M$ is $(|M|-1) |G:M|$. 
Something about this feels off, but I'm not sure; I never have any confidence in my proofs of combinatorial problems. I am especially uneasy about the second to last sentence. 
 A: If $G$ is infinite, then $|M|$ or $|G:M|$ is infinite, so $(|M|-1)|G:M|$ is infinite. Then we are done. (Not sure about this part.)
Let $G$ be finite. Then as you said, $M=N_{G}(M)$. Let $g_{1}Mg_{1}^{-1}=M,g_{2}Mg_{2}^{-1},\cdots,g_{k}Mg_{k}^{-1}$ be all the distinct conjugates of $M$. Let $A=\{M,g_{2}Mg_{2}^{-1},\cdots,g_{k}Mg_{k}^{-1}\}$. Let $G$ act on $A$ by conjugation: $g\cdot g_{i}Mg_{i}^{-1}=gg_{i}Mg_{i}^{-1}g^{-1}$. We can check that this is a group action. Since $g_{j}g_{i}^{-1}\cdot g_{i}Mg_{i}^{-1}=g_{j}Mg_{j}^{-1}$, the action is transitive. So there is only one orbit. By a proposition (e.g., Dummit and Foote page 114 proposition 2), $k=|A|=|G:G_{M}|=|G:N_{G}(M)|=|G:M|$. Now, 
$$
(\cup_{g\in G} gMg^{-1})-\{1\}=(\cup_{i=1}^{k} g_{i}Mg_{i}^{-1})-\{1\}=\cup_{i=1}^{k}(g_{i}Mg_{i}^{-1}-\{1\}).
$$
So 
$$
|(\cup_{g\in G} gMg^{-1})-\{1\}|=|\cup_{i=1}^{k}(g_{i}Mg_{i}^{-1}-\{1\})|\\
\leq \sum_{i=1}^{k} |g_{i}Mg_{i}^{-1}-\{1\}|=k(|M|-1)=(|M|-1)|G:M|.
$$
A: As you said $M=N_G(M)$.
Thus $[G:M]\,=\textit{number of conjugates of $M$ in }G$.
Each conjugate contains at most $(|M|-1)$ non identity elements in $G$, so the number of atmost nonidentity elements in $G$ that are contained in  conjugates of $M$ is equal to $(|M|-1|) [G:M]$
