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 Mersenne prime (?). Consider:

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

Then, any group of order $n$ has a nontrivial Normal Sylow subgroup (?): either the $2$-Sylow subgroup is normal, or the $(2^p - 1)$-Sylow subgroup is normal. Thus, $G$ admits a Sylow tower (?), so $G$ is a 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. congruence condition on Sylow numbers
2. 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.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 147, Section 4.5 (Sylow's theorem), Exercise 13, ^{More info}