 Sarkozy's theorem guarantees that if g is a bellion and a prime p divides the order of g, then g will have a subgroup of order p. But what if g is not a bellion, or p squared divides the order of g? For that, we'll need a stronger result. And in group theory, that stronger result is known as Selo's theorem, named after another Norwegian mathematician by the name of Selo. Suppose p is prime and p to the m is the highest power of p that divides the order of g, a bellion or not, then g has a subgroup of order p to power m. So remember a group is a set with a binary operation that meets certain requirements. So if we're looking for a subgroup, we might form a subgroup of order p to the m by choosing a set of p to the m elements. Generally speaking, this won't be a group, but hopefully we'll find one that is. So let the order of g be p to the m times k, where p is not a divisor of k, then there are p to the mk choose p to the m subsets with p to the m elements. And it's relatively straightforward to prove that p does not actually divide this binomial coefficient. That's a standard result we can obtain from combinatorial arguments. Now we could prove this here, but it would take us a little bit far afield from our proof of Selo's theorem, so we'll leave this as an exercise for the viewer. So let p be the set of all subsets of g with p to the m elements. p is a g set with our function from g cross p into p to find as follows we'll form the left coset of any element of p. Now we should actually establish two things. First of all, the image is going to be an element of p and also that this function is going to be a group action. So let's consider, consider any element in p and remember this element is actually a subset of g with p to the m elements. So a has p to the m elements and we're forming a left coset and since our group operation satisfies the Latin square property that means this left coset ga will also have p to the m elements. And so that means it's a set with p to the m elements and so ga is going to be an element of p. We also need to establish that this function is in fact a group action. So again we also know that if e in g is our identity then the coset we formed by left multiplication by the identity is just going to be the original set. And also for p and q in our group g we know that p applied to the coset qa will be pq applied to a. So let our set of sets be partitioned into orbits O1 through however many we have and since every element of p is going to be in exactly one of these orbits that means that the order of p is going to be the sum of the orders of the different orbits. Now since p does not divide the order of our set then there's at least one term in our sum that's not divisible by p and we'll just say that is the order of this orbit of some element s. Now here's where that orbit stabilizer theorem comes in. So if we have a g set and stabilizer then our orbit and stabilizer multiply to get the order of the group. And so the orbit stabilizer theorem tells us the order of this group is the order of the stabilizer times the cardinality of this particular orbit. But since our prime does not divide the order of the orbit then our prime and consequently our mth power of the prime must divide the order of the stabilizer. Now here's where things get a little tricky and the important thing to remember is what each of these sets consist of. So remember the stabilizer of an element s is the elements of our group where g applied to s gives us s. But the elements of p are actually subsets of g. So the stabilizer of s is actually the stabilizer of s where s is a subset of g. And that means we can use our theorem, let g be a group and k a subset of g. The stabilizer of the subset is a divisor of the cardinality of the subset. So that means the order of the stabilizer is a divisor of the cardinality of this subset and again all these subsets have cardinality p to the m. And so if we put everything together if the order of g is p to the mk then among the subsets of g with p to the m elements there's a subset s where the order of the stabilizer of s is divisible by p to the m. But the order of that stabilizer must also be a divisor of p to the m. And so that means this stabilizer has order p to the m and since it's a subgroup of g this proves Silo's theorem.