Results 1 to 7 of 7

Math Help - subgroups

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    34

    subgroups

    If G is an abelian group and n>1 an integer, let A_n = { a^n, a is an element of G}. Prove that A_n is a subgroup of G
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Godisgood View Post
    If G is an abelian group and n>1 an integer, let A_n = { a^n, a is an element of G}. Prove that A_n is a subgroup of G
    We use the one step subgroup test.

    First note that e=e^n, so A_n\neq\emptyset.

    Now, take a^n,b^n\in A_n. It follows then

    a^n\left(b^n\right)^{-1}=a^n\left(b^{-1}\right)^n=\left(ab^{-1}\right)^n.

    Since ab^{-1} is in the right form, it follows that A_n\leq G

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    34
    No not really.
    What I tried doing was showing that A_n is closed, associative, has an identity element and an inverse element.
    My problem was I got an identity element e = a^0
    and inverse = a^-n.. I suppose both of them are wrng because th question says n>1....what do you think I did wrong..
    Thnx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Godisgood View Post
    No not really.
    What I tried doing was showing that A_n is closed, associative, has an identity element and an inverse element.
    My problem was I got an identity element e = a^0
    and inverse = a^-n.. I suppose both of them are wrng because th question says n>1....what do you think I did wrong..
    Thnx
    To prove something is a subgroup you merely need to show Closure and that Inverses exist. You get associativity for free because you are in a group (think about it...), and the identity because inverses exist (if a and a^{-1} \in H and aa^{-1} \in H then as aa^{-1}=1 we have that 1 \in H).

    Firstly, note that your identity is e^0. Secondly, does the question mention "finite" anywhere?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    34
    No the question does not mention infinite..
    Alo bare with me but am still confused as to why the identity element is e^0
    I though a * e = a
    and in this case since A_n = a^n
    so is a ^n * e = a^n
    that is how I got e = a^0
    what am doing wrong
    Thanks for your help so far
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Godisgood View Post
    No the question does not mention infinite..
    Alo bare with me but am still confused as to why the identity element is e^0
    I though a * e = a
    and in this case since A_n = a^n
    so is a ^n * e = a^n
    that is how I got e = a^0
    what am doing wrong
    Thanks for your help so far
    The element a is any element of the group, while the n remains fixed.. For instance, take G=C_8=\{1,g,g^2, \ldots, g^7\}, with g^8=1. Let us look at this group when n=2, A_2. This is the set of squares of all elements of the group:

    A_2 = \{1^2=1, g^2, g^4, g^6 \}.

    The identity element is NOT a^0 - the n is fixed. It is the elements of the group that we change. The identity occurs when a=1 as 1^n=1 \forall n \in \mathbb{N}.

    If g \in G then g^{-1} \in G. Thus g^n \in A_n and (g^{-1})^n = g^{-n} \in A_n.

    For closure you need the abelian condition of G.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2009
    Posts
    34

    Smile

    I finally get it!!!
    Thanks swlabr
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Subgroups and Intersection of Normal Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 1st 2010, 09:12 PM
  2. subgroups and normal subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 19th 2010, 04:30 PM
  3. subgroups
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 10th 2010, 10:56 AM
  4. Subgroups and Normal Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: December 9th 2009, 09:36 AM
  5. Subgroups and normal subgroups
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: October 13th 2007, 05:35 PM

Search Tags


/mathhelpforum @mathhelpforum