Answer:
Explanation:
Let G be a group of order 143.
By the Sylow Third Theorem, the number of Sylow 11-subgroups of G is 1+11k and is a factor of 13,
Then, k=0 and there is an element a∈G such that Ord(a)=11 and ⟨a⟩ is normal to G.
Also, the number of Sylow 13-subgroups of G is 1+13k and is a factor of 11,
Then, k=0 and there is an element b∈G in such away that Ord(b)=13 and ⟨b⟩ is normal to G.
Since ⟨a⟩∩⟨b⟩={e}, it was discovered that ab=ba, therefore Ord(ab)=143.