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Proved characteristic subgroup without using all facts

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data
    This is question 30, section 2.5 from "Abstract Algebra 3rd edition" by Herstein.

    2. Relevant information
    A subgroup H of group G is called characteristic if for all automorphisms phi of G, phi(H) is a subset of H.

    (I paraphrased this from question 29. I don't know why he says "phi(H) is a subset of H" but not "phi(H) and H are the same set". After all, phi is bijective?)


    3. The attempt at a solution
    Here is what I did:
    H is a subgroup of order prime, so it is cyclic, generated by any non-identity element in H. Call H <a>.

    Let phi be an arbitrary automorphism on G, then phi must map <a> to yet another cyclic subgroup of order p in G, call this <phi(a)>. Since both <a> and <phi(a)> are cyclic, they're either the same subgroup, or only share the identity element.

    Let K of be the smallest subgroup generated by every possible products of <a> and <phi(a)>. If <a> and <phi(a)> are different, then the order of K is p2. So p2 has to divide pm, which contradicts the fact that p does not divide m. So <a> and <phi(a)> have to be the same subgroup in G.

    Hence any automorphism phi on G will also be an automorphism on <a>, which implies that <a>, aka H, is characteristic.


    My problem: I never used the fact that H is a normal subgroup. I'm being quite confused. Where was I wrong? Many thanks for the help.
     
  2. jcsd
  3. May 5, 2010 #2

    jbunniii

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    Why must the order of [itex]K[/itex] be [itex]p^2[/itex]? You haven't proved that. All that can be concluded based on what you have written so far is that if [itex]a[/itex] and [itex]\phi(a)[/itex] generate two different subgroups, then any subgroup containing them must have at least [itex]2p - 1[/itex] elements and its order must be divisible by [itex]p[/itex].
     
  4. May 5, 2010 #3

    Dick

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    jbunniii found your flaw. Now try to fix it. Try to show H is the ONLY normal subgroup of order p. Suppose there are two. Use that the product of normal subgroups is a group.
     
  5. May 8, 2010 #4
    Thanks Jbunniii and Dick for the help.

    I made my mistakes when concluding |K| = p2 too early. Here was how I came to it:
    Suppose ai phi(a)j = ak phi(a)l (where i, j, k, l are integers in range [0, p - 1])
    That implies ai-k phi(a)j-l = e
    Because we assumed <a> and <phi(a)> were different subgroups, each cannot contain the inverse of another, unless it is the identity element. This implies i = k and j = l.
    So a pair of (i, j) uniquely defines an element in <a>*<phi(a)>. There are p2 possible combinations.

    My mistake is I assumed <a>*<phi(a)> was already a group. It turned out I needed to use the fact that both <a> and <phi(a)> were normal subgroups to come to that conclusion. Thanks Dick for pointing that out.

    If my proof still contains any flaw, I'd appreciate any help. Thanks again :)
     
  6. May 8, 2010 #5

    Dick

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    Your proof is assuming that a and phi(a) commute when you rearranged the powers. That's not necessarily so. You should be able to show the product of two normal subgroups H and H' of order p has order p^2 without assuming that their generators commute.
     
  7. May 11, 2010 #6
    You should also be able to prove that dropping the assumption that H' is normal.

    Also section 2.5 in my copy of Herstein is headed "A Counting Principle". If the same is true in yours you can use it's main result

    [itex]o(HK)=\frac{o(H)o(K)}{o(H\cap K)}[/itex]

    without proving it again.
     
    Last edited: May 11, 2010
  8. May 11, 2010 #7
    Actually sorry to interfere, but where does phucnguyen's proof assume that a and phi(a) commute? With the fix you suggested it now looks OK to me.
     
    Last edited: May 11, 2010
  9. May 11, 2010 #8

    Dick

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    In post 4 phucnguyen assumes all elements of the group generated by the product can be written as a^i*phi(a)^j moving all of the factors of a to the left and all the factors of phi(a) to the right. That's assuming a and phi(a) commute.
     
  10. May 11, 2010 #9

    jbunniii

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    But HK isn't generally a group if neither H nor K is normal. And if HK is not a group, why does its order necessarily have to divide that of G (to obtain the contradiction)?
     
  11. May 12, 2010 #10
    The relevant word here is neither. We are given H is normal.
     
  12. May 12, 2010 #11
    No. It's just pre-multiplying both sides by a-k and post-multiplying both sides by phi(a)-l. This finds the number of elements in the set (a)(phi(a)) which as you pointed out is also a group because (a) and (phi(a)) are normal subgroups.

    That is, in post 4 phucnguyen appears to no longer start by considering the group generated by the product, but only the product itself as a set. Here the elements are defined with powers of a on the left and powers of phi(a) on the right.
     
    Last edited: May 12, 2010
  13. May 12, 2010 #12

    jbunniii

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    Oops, yes, you're right. I forgot to scroll back and re-read the problem statement.
     
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