Sylow-p Subgroups' Intersections With Normal Subgroups (3.3.9)

Dummit and Foote, section 3.3, exercise 9:

Let p be prime with group G, and | G | = p^am, where p does not divide m. Assume P is a Sylow-p subgroup in G, ie | P | = p^a and N is a normal subgroup with | N | = p^bn, where p does not divide n. Prove that P∩N is a Sylow-p subgroup in N and | PN/N | = p^a-b.

Proof: Since N is normal, PN ≤ G. Now by the Second Isomorphism Theorem:

PN/N ≅ P/(P∩N) ⇒
| PN/N | = | P/(P∩N) | ⇒
| PN : N | = | P : P∩N | = x, a placeholder.

If we draw the lattice diagram, we obtain the following:

Note that the fact that | PN : P | = y₁ = | N : P∩N | = y₂ follows from exercise 3.2.11, which states | PN : P∩N | = xy₁ and | PN : P∩N | = xy₂, thus y₁ = y₂ ; this is actually a general case of the Second Isomorphism Theorem.

Now, w is a power of p since P∩N is a subgroup of the Sylow-p subgroup P. As well, y clearly divides yz = | G : P | = (p^am)/p^a = m. Since p does not divide m, this implies p does not divide y. Since wy = | N | = p^bn, this can only mean that w = | P∩N | = p^b. Since wx = | P | = p^a, this means x = | PN/N | = p^a-b.

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