Results 1 to 9 of 9

Math Help - Normal Subgroup

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    13

    Normal Subgroup

    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by RoboMyster5 View Post
    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false. So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
    Yeah, that's the idea. You can use the fact that if G \cong A_4 \times C_3 and H \leq G then H \cong H_1 \times H_2 where H_1 \leq A_4 and H_2 \leq C_3. Essentially, that every subgroup of A_3 \times C_3 is a direct product of subgroups from these respective groups.

    However, you should probably look at your proof of A_4 not having a subgroup of order 6 as |A_4| = 12, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of [tex]A_4[\math] look like).

    Why is your thread called "normal subgroups"?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Whats C3? Thats the chapter the problem is in so I thought maybe somehow normal sbgroups was suppose to be involved.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Swlabr View Post
    Yeah, that's the idea. You can use the fact that if G \cong A_4 \times C_3 and H \leq G then H \cong H_1 \times H_2 where H_1 \leq A_4 and H_2 \leq C_3. Essentially, that every subgroup of A_3 \times C_3 is a direct product of subgroups from these respective groups.


    Why is this true in this case? This is definitively not so in other cases, for example: in the Klein group \mathbb{Z}_2 + \mathbb{Z}_2, the sbgp. <(1,1)>=\{(0,0),(1,1)\} is not the direct sum of sbgps. of \mathbb{Z}_2

    Tonio



    However, you should probably look at your proof of A_4 not having a subgroup of order 6 as |A_4| = 12, and 6 divides 12 (of course, I may be getting the wrong end of your stick here)...(if I was proving it, I'd look at what the elements of A_4 look like).

    Why is your thread called "normal subgroups"?
    .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by tonio View Post
    .
    hmm...good point, well made.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by RoboMyster5 View Post
    Prove that A4(+)Z3 has no subgroup of order 18.

    I know A4 does not have a subgroup of order 6 because the converse of Lagrange thereom is false.


    Well, that is not a reason but in fact a consequence: A_4 has no sbgp. of order 6 because it has none (Add proof here)...

    Here's an idea to prove what you want: suppose H\,,\,|H|=18 is a sbgp. of G=A_4\times C_3.

    Since |HA_4|=\frac{|H|\cdot |A_4|}{|H\cap A_4|}\le 36, we get that |H\cap A_4|\ge \frac{18\cdot 12}{36}=6\Longrightarrow The only option is |H\cap A_4|=6 (why??) , which contradicts that A_4 has no sbgp. of order 6 .

    Tonio


    So my initial notion is that because 6 is a multiple of 18, it cannot exist. Any thoughts on how to make this a proof?

    Thanks All.
    .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    Posts
    13
    I'm still not comprehending what C3 is, but I am following your idea with the orders.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by RoboMyster5 View Post
    I'm still not comprehending what C3 is, but I am following your idea with the orders.

    C_3 is the cyclic group of order 3. It is the same (or better, isomorphic to) as \mathbb{Z}_3, but the notation C_3 is used when we want to remark the group is a multiplicative one, as opposed to the additive \mathbb{Z}_3

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Oh okay now this makes a lot more sense, wow thank you so much.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 2nd 2011, 09:07 PM
  2. Subgroup of cyclic normal subgroup is normal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 25th 2010, 07:13 PM
  3. characterisitic subgroup implies normal subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 8th 2010, 04:13 PM
  4. subgroup of a normal subgroup...
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 23rd 2009, 09:06 AM
  5. Normal subgroup interset Sylow subgroup
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 10th 2008, 01:21 AM

Search Tags


/mathhelpforum @mathhelpforum