█ Q2. Let G be a group. Suppose that the number of elements in G of order 5 is 28. Determine the number of distinct subgroups of G of order 5.
▶ Solution. Let g be an element in G such that ∘(G) = 5.
Then the subgroup ⟨g⟩ generated by g is a cyclic group of order 5.
In other words, if e is the identity element in G, then ⟨g⟩ = {e, g, g2, g3, g4}.
Clearly, the order of each non-identity element in ⟨g⟩ is 5. Moreover, if h is another element in G of order 5, then either ⟨g⟩ = ⟨h⟩ or ⟨g⟩ ∩ ⟨h⟩ = {e}. It follows from the fact that the intersection ⟨g⟩ ∩ ⟨h⟩ is a subgroup of the order 5, and thus ∘(⟨g⟩ ∩ ⟨h⟩) is either 5 or 1.
Again, if H is a subgroup of G of order 5, then every non-identity element in H has order 5.
These observations imply that each subgroup of order 5 contains exactly 4 elements of order 5 and each element of order 5 appears in exactly one of such subgroups.
Since there are 28 elements of order 5, there are 28/4 = 7 subgroups of order 5. ▲