First suppose P is a Sylow p-subgroup of G. Then P∩H is a subgroup of P, so P∩H is a p-subgroup of H. Also P/P∩H≌PH/H, hence | H|/| P∩H| = | PH|/| P|. Furthermore | G|/| P| = | G|/| PH|·| PH|/| P| and we deduce that | H|/| P∩H| divides | G|/| P|. Since | G|/| P| is prime to p, it follows that | H|/| P∩H| is also prime to p, which proves that P∩H is a Sylow p-subgroup of H.
Next, PH/H≌P/P∩H, so PH/H is a p-subgroup of G/H. Furthermore | G|/| P| = | G|/| PH|·| PH|/| P|, and we see that | G|/| PH| is prime to p. Therefore | G/H|/| PH/H| is also prime to p and it follows that PH/H is a Sylow p-subgroup of G/H.
Now suppose P∩H and PH/H are Sylow p-subgroups. Since P/P∩H≌PH/H and | P| = | P/P∩H|·| P∩H|, we see that P is a p-group. Finally if | H| = pax and | G/H| = pby, where x and y are prime to p, then | G| = pa+bxy and xy is prime to p. This means that a Sylow p-subgroup of G has order pa+b. But since | P∩H| = pa and | PH/H| = pb, we see that | P| = pa+b and hence P is a Sylow p-subgroup of G, as required.
Now consider the Sylow 3-subgroups. If P and Q are distinct Sylow 3-subgroups and 1≠x∈P∩Q, then CG(x) contains P and Q and hence contains an element of order 2. It follows that G has an element of order 6, which is not possible by the previous paragraph, so distinct Sylow 3-subgroups intersect trivially.
Next the number of Sylow 3-subgroups is 16 or 64. If it is 16, we consider the representation of G on the left cosets of a Sylow 3-subgroup P. If Q is another Sylow 3-subgroup, then there is an orbit under Q which has order 3, which shows that there exists 1≠q∈Q such that q∈P∩Q, contradicting the previous paragraph. Finally if there are 64 Sylow 3-subgroups, then since two distinct Sylow 3-subgroups intersect in the identity, we can count elements to show that G has a normal Sylow 2-subgroup.
Now let I⊲R. We want to prove I is a principal ideal, and since 0 is clearly a principal ideal, we may assume that I≠ 0. Each nonzero element of I has a factorization upiqj, where u is a unit and i, j are nonnegative integers. Choose 0≠x∈I such that x = piqj, with i as small as possible, and then choose 0≠y∈I such that y = pkql, with l as small as possible. We will show that I = (piql). Clearly I⊆(piql). On the other hand pk-i + qj-l is a unit by the above, hence piql is an associate of y + x and we see that piql∈I. This proves that I = (piql).
Now we have well defined homomorphisms θ : ℤ/pℤ⊗ℤℤ/(| L|-- 1)ℤ and φ : ℤ/pℤ→ℤ/pℤ⊗ℤℤ/(| L|-- 1)ℤ determined by θ(⊗) = and φ() = ⊗, and θφ and φθ are the identity maps. This shows that ℤ/pℤ⊗ℤℤ/(| L|-- 1)ℤ≌ℤ/pℤ and we deduce that M≌(ℤ/pℤ)n. Therefore if (| K|,| L|-- 1)≠1, it follows that | M| = | K|.