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Sylow's Theorem

  1. Feb 7, 2008 #1
    Let p,q be distinct primes with q < p and let G be a finite group with |G| = pq.

    (i) Use sylow's theorem to show that G has a normal subgroup K with [itex]K \cong G [/itex]

    (ii) Use the Recogition Criterion to show [itex] G \cong C_p \rtimes_h C_q [/itex] for some homomorphism [itex] h:C_q \rightarrow Aut(C_p) [/itex]

    (iii) Describle explicitly all homomorphisms [itex] h:C_5 \rightarrow Aut(C_7) [/itex]. Hence describe all groups of order 35. How many such subgroups are there?

    (iv) Describe explicitly all homomorphisms [itex] h:C_3 \rightarrow Aut(C_{13}) [/itex]. Hence describe all groups of order 39. How many such groups are there, up to isomorphism?

    any help is highly appreciated as usual. i will attempt the rest myself once i have good idea. thnx a lot :)
     
  2. jcsd
  3. Feb 7, 2008 #2

    morphism

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    (i) is trivial as stated. Did you mean to say K=~C_p?
     
    Last edited: Feb 7, 2008
  4. Feb 11, 2008 #3
    no the first part was meant as it was shown. but do you have an idea for the other parts please? i don't have a strong enough idea right now to solve them but if you provide a clue then i'll attempt it myself. thnx a lot :)
     
  5. Feb 11, 2008 #4
    ok i managed to get part (i)

    G a Sylow q-subgroup, we let x be the number of these subgroups. As a result, x|p so x=1 or x=p and x = 1 (mod q) so x=1 or p.
    x cant be p because p = 1(mod q) which contradicts the fact that q>p. Therefore, x=1 and so its Sylow q-subgroup has to be a normal subgroup.

    please verify this.
    any help on the other parts would be greatly appreciated :)
     
  6. Feb 11, 2008 #5

    NateTG

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    You might as well write [itex]n_q[/itex] rather than [itex]x[/itex], and the argument could be clearer:
    Sylow gives:
    [tex]n_q | p \rightarrow n_q \in \{1,p\} \rightarrow n_q \leq p[/tex]
    then
    [tex]p < q \rightarrow n_q \leq q[/itex]
    thus
    [tex]n_q \equiv 1 (\rm {mod} q) \rightarrow n_q = 1[/tex]

    I'm unfamiliar with the expression 'recognition criterion' but the second part is similar to this problem -- I'm not sure if you got that one.
    https://www.physicsforums.com/showthread.php?t=213750
     
  7. Feb 12, 2008 #6
    for part (ii) i cant find any decent material to learn off and wikipedia doesnt seem to have much on the recognition theorem. any help on this is therefore appreciated aswell.

    for part (iv)
    could you help me on this part please? how can you show the homomorphisms explicitly?

    ok for this part this is what i got:
    h is a group of order 39 = 13 x 3.
    [itex]n_{13}[/itex] must divide 3. and [itex]n_{13} = 1(mod 13)[/itex]. The only value satisfying these constraints is 1. So there is only 1 subgroup of order 13.
    Similarly, [itex]n_3[/itex] must divide 13. and [itex]n_3 = 1(mod 3)[/itex]. The only value satisfying these constraints is 1. So there is only 1 subgroup of order 3.
    Now since 13 and 3 are co prime, the intersection of these 2 subgroups is trivial and thus there is only 1 group of order 39 upto isomorphism...

    is this done correct? please verify and correct if necessary?
     
  8. Feb 12, 2008 #7

    NateTG

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    "Recognition Criterion" not "Recognition Theorem".

    Regarding:
    [tex]C_3 \rightarrow Aut(C_{13})[/tex]
    Can you describe [itex]Aut(C_{13})[/itex]?
    -

    [itex]n_3=13[/itex] also is 1 mod 3, and divides 13...
     
  9. Feb 12, 2008 #8
    still cant manage to find much about that im afraid? any more ideas on that?

    ok so [itex]Aut(C_{13})[/itex] is:
    [itex]C_{13} = {1,x,x^2,x^3,x^4,..........,x^{12}}[/itex]

    so [itex]Aut(C_{13}) = {\phi_1,\phi_2,\phi_3,\phi_4,\phi_5,\phi_6,\phi_7,\phi_8,\phi_9,\phi_{10},\phi_{11},\phi_{12}}[/itex] right? so now what would be the next step?

    ok so if n_3 = 13 is also a possibility then there are only 2 subgroups of of order 3.. how does this change the final answer though?
     
    Last edited: Feb 12, 2008
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