 Solution

ID: past-exam-of-the-algebra-course-of-the-university-of-oxford/2011/6/a/ii/solution

S, T

t \neq = 1

, they are abelian by the preceding deduction, so

S, T \leq C_{G} (t)

. But

∣ C_{G} (t) ∣ = 8

and both subgroups have order eight. Thus

S = C_{G} (t) = T

. Consequently distinct Sylow $2$ -subgroups intersect only in the identity.

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