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\subsection*Exercise 15 Prove that there is no simple group of order $56 = 2^3\cdot 7$.

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A. Mouri's Repository : Another prominent set of solutions, though the author notes they are not a professional mathematician and some inaccuracies may exist. \subsection*Exercise 15 Prove that there is no simple

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While there isn't a single "official" essay on this topic, the intersection of Dummit & Foote’s Abstract Algebra , and the culture of shared solutions