Dirac's quantization rule I first recall Dirac's quantization rule, derived under the hypothesis that there would exit somewhere a magnetic charge: $\frac{gq}{4\pi} = \frac{n\hbar}{2} $ with $n$ a natural number.
I am wondering how the quantization of electric charge can be deduced from it. The quantization of the product $gq$ is certainly not enough; what else is demanded?
 A: i) First of all, the Dirac quantization rule 
$$\tag{1} \frac{qg}{2\pi\hbar} ~\in~ \mathbb{Z} $$
for magnetic monopoles can be generalized to the Dirac-Zwanziger-Schwinger quantization condition 
$$\tag{2} \frac{q_1g_2-q_2g_1}{2\pi\hbar} ~\in~ \mathbb{Z} $$
for dyons. (In a slight misuse of terminology, we shall in the following also include purely electrically charged particles and pure magnetic monopoles into the definition of dyons.) 
II) Let $\Gamma=\{(q,g)\}$ denote the set of electric and magnetic charges for dyons. It is natural to think of $\Gamma$ as a subset of the plane $\mathbb{R}^2$. The left-hand side of (2) has a geometric meaning as a signed area spanned by two vectors $(q_1,g_1)$ and $(q_2,g_2)$. 
III) Now assume that $\Gamma\backslash\{(0,0)\} $ is non-empty, i.e. there exists a dyon $(q_1,g_1)\neq(0,0)$ to begin with. What points $(q_2,g_2)\in\Gamma$ of $\mathbb{R}^2$ would not conflict with condition (2)? The answer is a set of equidistant discrete lines parallel to the vector $(q_1,g_1)$. 
IV) Now assume that $\Gamma$ contains at least two linearly independent vectors $(q_1,g_1)$ and $(q_2,g_2)$. What points $(q_3,g_3)\in\Gamma$ of $\mathbb{R}^2$ would not conflict with condition (2)? The answer is a discrete grid/lattice of intersection points, namely precisely where the corresponding two sets of equidistant discrete parallel lines from section III meet. In other words, the charges are quantized.  
V) As a special case, if there exist at least one purely electrically charged particle and at least one pure magnetic monopole, we are in the situation described in section IV, and hence the charges must be quantized.
A: I think this is a valid open question. If it turned out that there wasn't just one magnetic charge g, but a continuum of magnetic charge, then the quantization condition would not be a sufficient explanation for e. 
However to actually prove whether or not a continuum of magnetic charge leads to any contradiction would require at the very least solving an n-body problem with multiple magnetic charges, which is not a trivial  matter even for professionals in the field. (or at least a 3 body problem with 2 magnetic charges to see whether a contradiction or reaffirmation of the quantization arises)
If this subject has been touched upon in the literature it would be nice for someone with the knowledge to give some citations as reference material for those interested.
A: 1) Assume there exists a minimum nonzero electric charge, $q_0$. Therefore, the minimum magnetic charge is
$$ g_0 = \frac{2\pi}{q_0}. $$
2) Secondly, if the theory preserves C and CP. Then the dyon $(q,g_0)$ automatically implies a conjugate dyon $(-q,g_0)$. Applying Dirac-Zwanziger (see, in @Qmechanic answer) condition for these two dyons
$$ 2qg_0 = 2\pi n, $$
or,
$$ q = \frac{n}{2}q_0. $$
So we have two possibilities, the electric charge $q$ takes integer multiples of $q_0$, or takes odd integer multiples of $q_0/2$.
A: I will try to answer from a pure mathematical perspective. The quantization rule states that for any possible $q$ and $g$, there is some $n\in\mathbb{Z}$ such that $qg=nh$.
Now consider the set $X=\{n\in\mathbb Z^+|\exists q\in Q^+, g\in G^+\ \mathrm{s.t.} \ qg=nh\}$, where $Q^+$ contains all the possible positive charges and $G^+$ contains all the possible positive magnetic charges (of magnetic monopoles). Consider the minimal element $n_0$ in the set $X$, then there is some $q_0$ and $g_0$ that satisfies $q_0g_0=n_0h$.
Note that it is often assumed that $n_0=1$, but it is not needed in this proof.
Now consider some $g\neq g_0$. Then $q_0g=nh$ for some $n$. Since $n_0$ is minimal, we have $n>n_0$. Since both $n$ and $n_0$ are integers, we have $n=pn_0+r, 0\leq r<n_0, p\in\mathbb{Z}^+$. If $r\neq 0$, then $0<q_0(g-pg_0)=rh<nh$, contradicting that $n$ is minimal. Therefore $r=0$. and $g=pg_0$, so $g$ is quantized in the unit of $g_0$. Similarly, we can prove that $q$ must be multiples of $q_0$.
Mathematically speaking, the only assumptions used above are that if $q$ is a valid charge, then $-q$ is also a valid charge, and that if $q_1$ and $q_2$ are valid charges, then $q_1+q_2$ is a valid charge. The same for $g$. I think these assumptions should be pretty natural given the physical nature of $q$ and $g$.
