Order is product of Mersenne prime and one more implies normal Sylow subgroup

Statement
Suppose $$p$$ is a prime number such that $$2^p - 1$$ is a fact about::Mersenne prime. Consider:

$$n = 2^p(2^p - 1)$$.

Then, any group of order $$n$$ has a nontrivial fact about::normal Sylow subgroup: either the $$2$$-Sylow subgroup is normal, or the $$(2^p - 1)$$-Sylow subgroup is normal. Thus, $$G$$ admits a fact about::Sylow tower, so $$G$$ is a fact about::group having a Sylow tower.

Examples
Some examples for small Mersenne primes are as follows:


 * $$p = 2, 2^p - 1 = 3, n = 12$$: In a group of order $$12$$, either the $$2$$-Sylow subgroup or the $$3$$-Sylow subgroup is normal.
 * $$p = 3, 2^p - 1 = 7, n = 56$$, In a group of order $$56$$, either the $$2$$-Sylow subgroup or the $$7$$-Sylow subgroup is normal.
 * $$p = 5, 2^p - 1 = 31, n = 992$$: In a group of order $$992$$, either the $$2$$-Sylow subgroup or the $$31$$-Sylow subgroup is normal.

Related facts

 * Congruence condition on Sylow numbers
 * Prime divisor greater than Sylow index is Sylow-unique

Cannot pinpoint to any one prime
We can find groups of order $$2^p(2^p - 1)$$ where only the $$2$$-Sylow subgroup is normal, and we can find groups where only the other subgroup is normal. Here are examples of both:


 * Only the $$2$$-Sylow subgroup is normal: Consider the general affine group $$GA(1,2^p)$$. This is the semidirect product of the additive group of a field of order $$2^p$$ with its multiplicative group. Here, only the $$2$$-Sylow subgroup is normal.
 * Only the $$(2^p - 1)$$-Sylow subgroup is normal: Consider the direct product of the dihedral group of order $$2(2^p - 1)$$ and a cyclic group of order $$2^{p-1}$$. Here, only the $$(2^p - 1)$$-Sylow subgroup is normal.

Facts used

 * 1) uses::congruence condition on Sylow numbers
 * 2) uses::divisibility condition on Sylow numbers

Proof
Given: A Mersenne prime $$2^p - 1$$, a group $$G$$ of order $$2^p(2^p - 1)$$.

To prove: Either the $$2$$-Sylow subgroup or the $$(2^p - 1)$$-Sylow subgroup of $$G$$ is normal.

Proof: Let $$q = 2^p - 1$$ and let $$n_q$$ and $$n_2$$ denote the number of $$q$$-Sylow subgroups and $$2$$-Sylow subgroups respectively. We have:

$$n_q \equiv 1 \mod q$$.

Also, we have, since $$n_q$$ equals the index of the normalizer of a $$q$$-Sylow subgroup:

$$n_q | 2^p$$

This forces either $$n_q = 1$$ or $$n_q = 2^p$$. If $$n_q = 1$$, we are done. Consider the case $$n_q = 2^p$$. Let $$Q_1, Q_2, \dots, Q_{2^p}$$ be all the $$q$$-Sylow subgroups. Since these are all groups of prime order, any two of them intersect trivially. Thus, the total number of non-identity elements in the subgroups is:

$$2^p(q - 1) = 2^p(2^p - 2) = n - 2^p$$.

Thus, there are exactly $$2^p$$ elements that are not among the non-identity elements of $$q$$-Sylow subgroups. Note that any element in a $$2$$-Sylow subgroup cannot be among these non-identity elements of $$q$$-Sylow subgroups, so any element in a $$2$$-Sylow subgroup must be among these $$2^p$$ elements. This forces any $$2$$-Sylow subgroup to be contained in this set of $$2^p$$ elements. Since the order of a $$2$$-Sylow subgroup is $$2^p$$, this forces there to be a unique $$2$$-Sylow subgroup -- precisely those $$2^p$$ elements. This completes the proof.