Results 1 to 3 of 3

Thread: Normal subgroup

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Normal subgroup

    Let G be a group, and let $\displaystyle H \leq G$

    (That means H is the normal subgroup of G, right?)

    a)Show that $\displaystyle aHa^{-1}$ is also a subgroup of G.

    My proof:

    Suppose that x,y is in G, so that $\displaystyle xHx^{-1}, yHy^{-1} \in aHa^{-1}$

    Since H is normal in G, pick $\displaystyle Hx^{-1}=x_{1}H, yH=Hy_{1},x_{1},y_{1} \in G$

    $\displaystyle (xHx^{-1})(yHy^{-1}) = x(x_{1}H)(Hy_{1}y^{-1}) = xx_{1}Hy_{1}y^{-1}$

    Now, $\displaystyle (xHx^{-1})^{-1}=x^{-1}Hx \in aHa^{-1}$

    Is this right?

    b) If H is finite of order k, then $\displaystyle aHa^{-1}$ is also finite of order k.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    Do you know this theorem?
    A set H in a group G is a subgroup iff $\displaystyle a \in H\;\& \;b \in H \Rightarrow \quad ab^{ - 1} \in H$?

    Suppose that $\displaystyle x \in aHa^{ - 1} \;\& \;y \in aHa^{ - 1} \Rightarrow \left( {\exists c \in H} \right)\left( {\exists d \in H} \right)\left[ {x = aca^{ - 1} \wedge y = ada^{ - 1} } \right]$.

    What can be said about $\displaystyle xy^{ - 1}$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tttcomrader View Post

    b) If H is finite of order k, then $\displaystyle aHa^{-1}$ is also finite of order k.
    $\displaystyle \{ a_1, ... , a_n \}$
    Let the above be the elements of $\displaystyle H$.
    Then,
    $\displaystyle \{aa_1a^{-1},...,aa_na^{-1} \}$
    Are all distint.
    Because otherwise if,
    $\displaystyle aa_ia^{-1} = aa_ja^{-1}$
    This implies,
    $\displaystyle a_i = a_j$.
    Which is an impossibility for distinct elements.
    -----
    Here is a useful theorem to know.

    Theorem: Let $\displaystyle H$ be a subgroup of $\displaystyle G$ the following are equivalent.
    1)For all $\displaystyle g\in G$ and $\displaystyle h\in H$ implies $\displaystyle ghg^{-1} \in H$.
    2)For all $\displaystyle g\in G$ we have $\displaystyle gHg^{-1}$.
    3)The left and right cosets coincide, i.e. $\displaystyle aH = Ha$.

    All these 3 are equivalent formulations for "normal subgroup". My favorite is #1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Mar 2nd 2011, 08:07 PM
  2. Subgroup of cyclic normal subgroup is normal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 25th 2010, 06:13 PM
  3. characterisitic subgroup implies normal subgroup
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Apr 8th 2010, 03:13 PM
  4. subgroup of a normal subgroup...
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Nov 23rd 2009, 08:06 AM
  5. Normal subgroup interset Sylow subgroup
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 10th 2008, 12:21 AM

Search Tags


/mathhelpforum @mathhelpforum