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Let G be a group of order 143 = 11 × 13, and, as usual, let Z(G) denote the center of G. Assume that we have found an element x ∈ Z(G) with x = e. What are the possibilities for the |Z(G)|? Prove any assertions you make.

User Mckelvin
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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.

User Jonathan Wilson
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