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

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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

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