Proving 1 p-Sylow Subgroup of G is Normal

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SUMMARY

If a group G has only one p-Sylow subgroup, that subgroup is normal. This conclusion arises from the fact that the unique p-Sylow subgroup S must be invariant under conjugation, meaning for any element g in G, the conjugate gSg^{-1} equals S. Therefore, the left cosets and right cosets coincide, confirming the normality of S within G.

PREREQUISITES
  • Understanding of group theory concepts, specifically Sylow theorems.
  • Familiarity with the definitions of normal subgroups and conjugation.
  • Knowledge of group operations and cosets.
  • Basic proficiency in mathematical proof techniques.
NEXT STEPS
  • Study the Sylow theorems in detail, focusing on their implications in group theory.
  • Learn about normal subgroups and their properties in finite groups.
  • Explore examples of groups with unique p-Sylow subgroups to solidify understanding.
  • Investigate the role of conjugacy classes in group theory.
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Mathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the properties of Sylow subgroups.

dogma
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Hello!

For the life of me, I can't seem to figure this out (vapor lock in the ol' brain):

Show that if G has only 1 p-Sylow subgroup, then it must be normal.

I know it something to do with showing it's a conjugate to itself (right coset = left coset?). I'm just not quite sure how to go about showing this.

Thanks for your time, help, and patience with me.

dogma
 
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Let S be the sylow subgroup. gSg^{-1} is another sylow subgroup, and must be S as S is unique.
 
thanks

Thank you...that makes sense.

dogma
 

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