# Extending Cauchy’s Theorem

Here I present an extension of Cauchy’s theorem that I haven’t seen anywhere else.

In our earlier proof of Cauchy’s theorem, we saw that r, the number of p-tuples (g,g,…,g) of elements of G whose product is e, had to be a multiple of p. Since (e,e,…,e) was one such p-tuple, we knew that r was greater than zero, and therefore concluded that there must be at least one other element g in G such that gp = e. And that was how we got Cauchy’s theorem. But in our final step, we weakened our state of knowledge quite a bit, from “r = pn (for some positive n)” to “r > 1”. We can get a slightly stronger result than Cauchy’s theorem by just sticking to our original statement and not weakening it.

So, we know that r = pn for some n. Does this mean that there are pn elements of order p? Not quite. One of these p-tuples is just (e,e,…,e), and e is order 1, not p. So there are really (pn – 1) elements of order p.

Furthermore, each of these elements forms a subgroup of size p. Every non-identity element in any of these subgroups is also order p. So this tells us that the number of elements of order p must be k(p – 1), where k is the number of subgroups of order p.

Putting these together, we see that (pn – 1) = k(p – 1). Crucially, this equation can’t be satisfied for all n and p! In particular, for k to be an integer, the value of n must be such that pn – 1 is divisible by p – 1. Let’s look at some examples.

p = 2

2n – 1 must be divisible by 1. This is true for all n.
So k, the number of subgroups of order 2, is 2n – 1 for any positive n.
k = 1, 3, 5, 7, …

p = 3

3n – 1 must be divisible by 2. This is only true for odd n.
So k, the number of subgroups of order 3, is (3n – 1)/2 for any odd n.
k = 1, 4, 7, 10, …

p = 5

5n – 1 must be divisible by 4. This is only true for n = 1, 5, 9, 14, …
So k = 1, 6, 11, 16, …

See the pattern? In general, the number of subgroups of order p can only be 1 + mp for any m ≥ 0. And the number of elements of order p is therefore mp2 – (m – 1)p – 1.

Needless to say, this is a much stronger result than what Cauchy’s theorem tells us!

## An Application

Say we have a group G such that |G| = 15 = 3⋅5. By Cauchy, we know that there’s at least one subgroup of size 3 and one of size 5. But now we can do better than that! In particular we know that:

Number of subgroups of size 3 = 1, 4, 7, 10, …
Number of subgroups of size 5 = 1, 6, 11, 16, …

For each subgroup of size 3, we have 2 unique elements of order 3. And for each subgroup of size 5, we have 4 unique elements of order 5.

Number of elements of order 3 = 2, 8, 14, …
Number of elements of order 5 = 4, 24, 44, 64, …

But keep in mind, we only have 15 elements to work with! This immediately rules out a bunch of the possibilities:

Number of subgroups of size 3 = 1, 4, or 7
Number of subgroups of size 5 = 1

So we know that there is exactly one subgroup of size 5, which means that 4 of our 15 elements are order 5. This leaves us with only 10 non-identity elements left, ruling out 7 as a possible number of subgroups of size 3. So finally, we get:

Number of subgroups of size 3 = 1 or 4
Number of subgroups of size 5 = 1

This is as far as we can go using only our extended Cauchy theorem. However, we can actually go a little further using Sylow’s Third Theorem. This allows us to rule out there being four subgroups of size 3 (since 4 doesn’t divide 5). So the “subgroup profile” of G is totally clear: G has one subgroup of size 3 and one of size 5. You can use this fact to show that there is exactly one group of size 15, and it is just 15.

15 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
Subgroup of size 3 = <5> = {0, 5, 10}
Subgroup of size 5 = <3> = {0, 3, 6, 9, 12}
All other elements generate the whole group.