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I Normalizer of a subgroup of prime index

  1. Oct 30, 2016 #1
    Hello! Can anyone help me with this problem?

    If H is a subgroup of prime index in a finite group G, show that either N(H)=G or N(H) = H.

    Thank you!
     
  2. jcsd
  3. Oct 30, 2016 #2

    fresh_42

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    What do you know about the index of a subgroup? And where is ##N_G(H)## settled with respect to ##H## and ##G##?
     
  4. Oct 30, 2016 #3
    Well as H has prime index it means it is cyclic. Also we have that p=ord(H) | ord(N) | ord(G). So we should have ord(N)=##k_1##p and ord(G)=##k_2##ord(N)=##k_1## ##k_2## p. So we have to show that either ##k_1##=1 or ##k_2##=1. This is what I got by now. But I don't know what to do next
     
  5. Oct 30, 2016 #4

    fresh_42

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    So p=ord(H) | ord(N) | ord(G) means ##H \leq N_G(H) \leq G##. The index of ##H## is also equal to ##|G/H|##. Do you know what happens, if you factorize ##H## across these inclusions?
     
  6. Oct 30, 2016 #5
    Sorry I am a little confused. You mean something like ##|H/H| \le |N(H)/H| \le |G/H| ##? This means ##1 \le |N(H)/H| \le |G/H|##. So what next?
     
  7. Oct 30, 2016 #6

    fresh_42

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    If ##|G/H|=p## prime, how many orders fit in between ##1## and ##p##?

    The essential part to be confused of, is the fact, that these cosets are possible disregarding whether ##H## is normal or not. Do you know why?
     
  8. Oct 30, 2016 #7
    Thank you so much! I kept reading a subgroup of prime order. It is index not order. Now it is easy. Sorry for this and thank you again!
     
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