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Sylow theorem

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Let G be a group of order 48. By the 1st Sylow theorem G has a Sylow 2-subgroup and a Sylow 3-subgroup. Suppose none of these are normal. Determine the number of Sylow 2-subgroups and Sylow 3-subgroups that G can have. Justify your answer.

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Solution Summary

Determine the number of Sylow 2-subgroups and Sylow 3-subgroups that group G can have.

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We note that |G| = 48 = 2^4 * 3
If m is the number of Sylow 2-subgroups that G may have, then m = 1 (mod 2). So m = 2k+1. We also have m|3, then ...

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