Sylow's Theorem Proof for Group of Order 35^3 | Validity Check

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SUMMARY

Any group of order 353 has a normal subgroup of order 125, as established through Sylow's theorems. Specifically, by Sylow's first theorem, a Sylow 5-subgroup of order 125 exists within the group. Sylow's second theorem indicates that this subgroup is conjugate to itself, leading to the conclusion that it is normal in the group. To confirm the normality of this subgroup, Sylow's third theorem can be applied to demonstrate that it is the only Sylow 5-subgroup present.

PREREQUISITES
  • Sylow's Theorems (First, Second, and Third)
  • Group Theory fundamentals
  • Understanding of normal subgroups
  • Basic knowledge of group orders and their implications
NEXT STEPS
  • Study the implications of Sylow's First Theorem in group theory
  • Explore Sylow's Second Theorem and its applications
  • Learn about Sylow's Third Theorem and its role in subgroup classification
  • Investigate examples of groups of order 353 to see these theorems in action
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, group theorists, and anyone interested in the application of Sylow's theorems in group structure analysis.

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Homework Statement



Show that any group of order 35^3 has a normal subgroup of order 125.


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The Attempt at a Solution



Is this a valid proof?


Let G be an arbitrary group of order 353. Note that 353 = 5373. Thus, by Sylow's first theorem, there is a sylow p-subgroup of order 125, which we refer to as H. But then, by Sylow's second theorem, it follows that H is conjugate to itself in G. Hence, H is normal in G.
 
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I believe you are on the right track, however, I believe you have made an assumption. Sylow's second theorem says that all the Sylow 5-subgroups are conjugate to each other. This does imply normality of H iff H is the only Sylow 5-subgroup of G. I think you can use Sylow's third theorem to show that H is the only Sylow 5-subgroup.
 

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