Answer :
Final answer:
By applying the Sylow Theorems we establish that any group of order 147 has a unique Sylow 7-subgroup. The uniqueness of any Sylow p-subgroup means it is normal, proving that every group of order 147 has a normal Sylow 7-subgroup.
Explanation:
The question asked is about proving that every group of order 147 has a normal Sylow 7-subgroup. This is about group theory in abstract algebra, part of higher level mathematics. According to the Sylow Theorems, for a finite group G of order 147 (which factors as 3*7^2), the number of Sylow p-subgroups (let's say p=7, so these are 7-subgroups) must be equivalent to 1 modulo 7 and at the same time it should divide 3. The only number satisfying this criterium is 1, meaning there is only one Sylow 7-subgroup.
When there is only one Sylow p-subgroup, it is normal in G (based on Sylow's theorem), so every group of order 147 has a normal Sylow 7-subgroup.
Learn more about Sylow Theorems here:
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