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Finite subgroups of unique orders are normal

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Let G be a finite group and H a subgroup of G having order m. Show that if H is the only subgroup of order m in G, then H is normal in G.


    2. Relevant equations
    A subgroup [itex]H[/itex] of [itex]G[/itex] is normal in [itex]G[/itex] if and only if [itex]xHx^{-1} \subseteq H \forall x \in G[/itex]


    3. The attempt at a solution
    Suppose that H is the only subgroup of order m. Then elements in G\H cannot have order m.

    If [itex]x \in H[/itex] then clearly, [itex]xHx^{-1} \subseteq H[/itex]

    If [itex]x \notin H[/itex] that is, [itex]x \in G \backslash H[/itex] so [itex]|x| \neq m[/itex] then ...

    This is where I'm not seeing anything. Any help to point me in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Nov 15, 2009 #2

    Dick

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    Is xHx^(-1) a subgroup? How many elements does it have?
     
  4. Nov 15, 2009 #3
    Since H is a subgroup of order m, xHx^(-1) will be a subgroup of order m. Since H is the only subgroup of order m there will be no other subgroup such that xKx^(-1) is of order m. Therefore, xHx^(-1) = H.

    I think that's it, but I can't connect the dots to the last statement.
     
  5. Nov 15, 2009 #4

    Dick

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    I don't see the problem. If xHx^(-1) is a subgroup of order m, and H is the ONLY subgroup of order m, then xHx^(-1) must be the same subgroup as H. What's K got to do with it?
     
  6. Nov 15, 2009 #5
    Oh wow, I swear I miss the most obvious things. Thanks for your help again!
     
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