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I Understanding a Sylow theory proof

  1. Dec 8, 2018 #1
    https://imgur.com/a/oSioYel

    I am trying to understand this proof, but am tripped up on the part that says "Consider the action of ##G## on ##\operatorname{Syl}_2(G)## by conjugation." My question is, how is this a well-defined action if ##\operatorname{Syl}_2(G)## is not normal? Isn't this action by conjugation defined only on subgroups that are normal?
     
  2. jcsd
  3. Dec 8, 2018 #2

    fresh_42

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    Given any operation ##\varphi \, : \, U\times G \longrightarrow G## with a subgroup ##U\leqslant G## we can define cosets ##g.U=\varphi(g,U)=\{\,g.u\,|\,u\in U\,\}## which in our case are the sets ##gUg^{-1}##. This partitions ##G## into subsets of equal size.

    Now the point is, that those sets are just that: sets. Since ##U## isn't normal, we simply cannot define a group structure on this set ##G/U## of sets. So as long as we are only counting elements, everything will be fine. If we want to consider homomorphisms ##U \rightarrowtail G \twoheadrightarrow G/U## we will have a problem if ##U## isn't normal. So normality directly corresponds to the fact, that ##G/U## is again a group or not.
     
  4. Dec 8, 2018 #3

    Infrared

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    [itex]Syl_2(G)[/itex] is not a subgroup of [itex]G[/itex]; rather, it is the set of all of the 2-Sylow subgroups of [itex]G[/itex]. If [itex]H\in Syl_2(G)[/itex], then [itex]gHg^{-1}\in Syl_2(G)[/itex] too (and can be a different element if [itex]H[/itex] is not normal).
     
  5. Dec 8, 2018 #4
    Got it. One more question. Why does that fact that ##\rho## is transitive imply that it is nontrivial?
     
  6. Dec 8, 2018 #5

    Infrared

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    Let [itex]H_1,H_2\in Syl_2(G)[/itex]. Since the action is transitive, there is an element [itex]g\in G[/itex] such that [itex]gH_1g^{-1}=H_2[/itex]. We can pick [itex]H_1,H_2[/itex] to be different, and then this means that [itex]\rho(g)[/itex] is not the trivial permutation.
     
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