= Solution
Put $n=|\Omega|\geq2$, and let $H=G_\alpha$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ <stabilizer subgroups> shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a <normal subgroup>, rather than presuming that fixed-point-free elements are closed under multiplication.
For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex <characters of a finite group>. Let $\pi$ be the <permutation character> and $\psi=\pi-1$, the character of the <permutation representation> with its constant line removed. For every nontrivial <irreducible character> $\theta$ of $H$, form the <virtual character>
$$
\varphi_\theta=\operatorname{Ind}_H^G\theta-\theta(1)\psi.
$$
Its values are $\theta(1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus\{1\}$; induction gives $\varphi_\theta(g)=\theta(h)$, because there is exactly one fixed coset.
The identity and the $n-1$ fixed-point-free elements together contribute $n\theta(1)\overline{\eta(1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ <stabilizer subgroups>. Hence <character orthogonality> gives
$$
\langle\varphi_\theta,\varphi_\eta\rangle_G
=\frac{n}{|G|}\sum_{h\in H}\theta(h)\overline{\eta(h)}
=\delta_{\theta,\eta}.
$$
A <virtual character> of norm one is plus or minus an <irreducible character>: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta(1)$ selects the plus sign. Thus each $\varphi_\theta$ is an actual <irreducible character>.
For a <group representation>, $\chi(g)=\chi(1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the $\varphi_\theta$. Conversely, a nonidentity element fixing a point gives $h\ne1$ in $H$. Some nontrivial <irreducible character> of $H$ has $\theta(h)\ne\theta(1)$, since otherwise the <regular representation> of $H$ would not vanish at $h$. Consequently
$$
N=\bigcap_{\theta\ne1_H}\ker\varphi_\theta.
$$
This proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a <regular permutation subgroup>.
Now $H$ acts transitively by conjugation on $N\setminus\{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is <elementary abelian> of order $p^d=n$.
Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every <group automorphism>. We have proved
$$
\boxed{N\text{ is regular, elementary abelian and characteristic in }G.}
$$
The character argument supplies the <regular kernel of a finite sharply two-transitive group>; the final Sylow argument establishes the stronger characteristic assertion.
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