Results 1 to 7 of 7

Math Help - Isomorphism Theorems

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    171

    Isomorphism Theorems

    The problem is:
    Let G be a group of order 12 ( o(G)=12).
    Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ).

    In the previous parts of the questions i've proved that:
    1a. a has 2 different conjucates in G and o(N(a))=6 or o(N(a))=12 (N(a) is the normalizer of a)
    1b. there is a b in G of order 2 who is commutative with a: ab=ba
    1c. if o(N(a))=6 then N(a) is a group of order 6 that is generated by ab ( <ab>=N(a) )
    1d. Prove that b is in the center of the group G (C(G)) . There's a clue: notice that b is in
    N(a).

    I don't under stand how to prove this one:
    2. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)...
    We know that G has no normal sub groups of order 3... We know that G is isomorphic to a subgroup of S4...But how can we know that the image is A4?

    Help is needed!
    TNX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by WannaBe View Post
    The problem is:
    Let G be a group of order 12 ( o(G)=12).
    Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ).

    In the previous parts of the questions i've proved that:
    1a. a has 2 different conjucates in G and o(N(a))=6 or o(N(a))=12 (N(a) is the normalizer of a)
    1b. there is a b in G of order 2 who is commutative with a: ab=ba
    1c. if o(N(a))=6 then N(a) is a group of order 6 that is generated by ab ( <ab>=N(a) )
    1d. Prove that b is in the center of the group G (C(G)) . There's a clue: notice that b is in
    N(a).

    I don't under stand how to prove this one:
    2. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)...
    We know that G has no normal sub groups of order 3... We know that G is isomorphic to a subgroup of S4...But how can we know that the image is A4?

    Help is needed!
    TNX

    I'm confused: at the beginning it is given that "Let's assume that G has a normal sub-group of order 3 and let a be her generator ( <a>=G ).", but this is absurd: I think they meant a is the generator of the normal sbgp. of order 3, not of G.
    Also, in (2), whatever the conditions are, it can't be G\cong A_4 , since A_4 has no normal sbgp. of order 3, as given at the beginning...check this and write back with al the conditions clearluy stated.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    171
    1 and 2 have different assumptions!!!
    The only thing we know in 2 is that 2 doesn't divide o(C(G)) and that o(C(G))=12 ...That's it! We use 1 only to get a contradiction when we assume G has a normal sbgrp of order 3... !!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by WannaBe View Post
    1 and 2 have different assumptions!!!
    The only thing we know in 2 is that 2 doesn't divide o(C(G)) and that o(C(G))=12 ...That's it! We use 1 only to get a contradiction when we assume G has a normal sbgrp of order 3... !!!

    Ok, that could have been said at the beginning... , but still: are we seeing G as embedded in some bigger group, or else what is C(G)?? Or is this a (non-standard) way to symbolize the center of the group G? Usually, C(G) means the centralizer of the group G inside a bigger group...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    171
    It's indeed the center of the group G...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by WannaBe View Post
    It's indeed the center of the group G...

    Ok, but then I'm still a little baffled...let's see what we've here: G is a group of order 12 s.t. 2 \nmid Z(G)...and that's all we have, right? But then you added:
    "2. prove that if 2 doesn't divide o(C(G)) then G is isomorphic to A4 (A4 is the group of all even permutations)...
    We know that G has no normal sub groups of order 3... We know that G is isomorphic to a subgroup of S4...But how can we know that the image is A4?"

    How do "we know" G has no normal sbgp. of order 3 if we haven't received more information?? There are groups of order 12 with normal sbgps. of order 3 (all the abelian ones and one of the non-abelian ones), and if we already know then we're almost done, since 12 = 2^2\cdot 3, and a simple counting argument shows that either the Sylow 2-sbgp. or the Sylow 3-sbgp. MUST be normal, so if it isn't the 3-sbgp. then it must be the 2-sbgp. of order 4 ==> G=A_4 as it is the only non-abelian group of order 12 with normal sbgp. of order 4...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    171
    I'll explain:
    G has order 12... By the first part, we know that if there's a normal sub-group of order 3, there's an element in Z(G) of order 2, and that's a contradiction to lagrange's theorem... So we now know there's no normal sbgrps of order 3... that means that all the sbgrps are from order different from 3... We can defind a homo. that makes use of these facts...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. isomorphism theorems
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 14th 2011, 03:16 PM
  2. Replies: 4
    Last Post: February 14th 2010, 03:05 AM
  3. Value Theorems
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 15th 2009, 10:42 PM
  4. Theorems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 4th 2007, 08:03 PM
  5. Theorems
    Posted in the Algebra Forum
    Replies: 5
    Last Post: July 15th 2006, 08:43 AM

Search Tags


/mathhelpforum @mathhelpforum