Actions by p-Groups (Pt. II)
Point of Post: This is a continuation of this post.
Theorem (Cauchy): Let be any group (not necessarily a -group) and a prime with . Then, there exists with .
Proof: Let . Since we can clearly choose the first coordinates of a tuple in and the last is determined as the inverse of the product of the first we have that . It can be easily verified that acts on via cyclic shifting (forward) of the coordinates of a tuple in . More explicitly,
Indeed, it clearly suffices to prove that for all since the other properties of a group action are apparent. But, this is clear by conjugation, namely
but, the very last expression is the product left-to-right product of and so as desired. Note then that by the fundamental theorem we have that is divisible by (since since ). But, since we must have that and so there exists with . That said, note that since is fixed by every element of we have that and so, in particular, and by construction . But we see then that since we have that . It follows then that and since we have that as desired.
From where we get the following, fairly relevant corollary:
Theorem: A finite group is a -group if and only if is a power of for every .
Proof: Evidently from Lagrange’s theorem we have that if is a -group then every element of has prime power order. Conversely, evidently if are two different primes dividing then by Cauchy’s Theorem we know that there exists with and . It follows then that every element of has order equal to the power of a single prime.
And a simple proof that the only finite fields can be of order for some prime . Indeed:
Theorem: Let be a finite field, then for some prime .
Proof: Note that the map is an (additive) isomorphism for each and so for all . Note though that if is divisible by two primes, we could find two elements of different orders. Thus, is a primer power as desired.
What we now wish to do is prove that if is a -group sitting inside the larger group . From where we will be able to prove that if then .
Theorem: Let be any finite group and a -subgroup. Then .
Proof: Consider the action of on given by left multiplication. What we’d like to figure out is what really is. Well, note that if then for all and so for some and so . Thus, since was arbitrary we may conclude that and so and conversely one can check that if then . Thus, . Thus, . The conclusion then follows from the fundamental theorem.
Corollary: If then , and since we may conclude that and so .
This tells us that if and has then , since by the corollary we have that and so clearly . That said, we already knew this from the theorem that says if the index of a subgroup is the smallest prime dividing the group then it’s normal. What this theorem is really useful for is:
Theorem: Let be a finite -group with . Then, there exists a chain of subgroups
Proof: We first use Cauchy’s Theorem to prove that there exists some with , we then take . We then proceed by ‘induction’. Suppose that we have found a group with and . We see then that and so by the previous corollary we have that . We see then that and so by Cauchy’s theorem we have that there exists with . But, from the correspondence theorem we know that for some and moreover that and so . To see that we may merely appeal to the paragraph immediately preceding this theorem. Thus, the ‘induction’ is complete.
This allows us to conclude from the first (weak) Sylow’s Theorem, namely the existence of Sylow -subgroups, the stronger result that if where is any group that contains a subgroup with .
To finish this post let’s give a cool proof of Wilson’s theorem:
Theorem(Wilson): Let . Then, if and only if is prime.
Proof: Suppose that is prime. Consider the action of on as previously discussed. We noted that is equipotent to and so by the Fundamental Theorem we have that . Note that if is non-identity then it’s order is the least common multiple of the cycle lengths of its disjoint cycle decomposition. It clearly follows that the only non-identity elements of of order are the -cycles. Now, how may distinct -cycles are there? Well, there are ways to arrange , but each of these arrangements is equivalent to precisely others (the cyclic shifts forward) and so the number of distinct -cycles is . Thus, since the only for which are the elements of order and we may conclude that , thus as previously noted .
Conversely, if has a proper divisor then we have to have that but by assumption it would also divide and so , which is a contradiction.