Results 1 to 5 of 5

Thread: Quotient Groups

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    37

    Quotient Groups

    Let H be a normal subgroup of finite group G. If the order of the quotient groups $\displaystyle G/H$ is m, prove that $\displaystyle g^m$ is in H for all $\displaystyle g \in G$.

    So since H is normal, there are no distinctions between left and right cosets in G.

    G is a finite group. Let o(G) = n. The o(G/H) = m.

    Don't know where to go after this...
    Last edited by MissMousey; May 4th 2010 at 03:22 PM. Reason: Using proper notation
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2007
    Posts
    70
    Hint: Given any group $\displaystyle G$ of order $\displaystyle n$ and any $\displaystyle g \in G$, $\displaystyle g^n = e$.

    In the case of $\displaystyle G/H$, what does this tell you about the coset $\displaystyle g^mH$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    37
    Quote Originally Posted by spoon737 View Post
    Hint: Given any group $\displaystyle G$ of order $\displaystyle n$ and any $\displaystyle g \in G$, $\displaystyle g^n = e$.

    In the case of $\displaystyle G/H$, what does this tell you about the coset $\displaystyle g^mH$?
    Since $\displaystyle G/H$ is of finite order m, does that imply that $\displaystyle g^mH = eH = H$?

    But if $\displaystyle g^mH = H$ then $\displaystyle g^m$ must $\displaystyle \in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $\displaystyle h \in H$).
    Last edited by MissMousey; May 4th 2010 at 02:40 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    Quote Originally Posted by MissMousey View Post
    Since $\displaystyle G/H$ is of finite order m, does that imply that $\displaystyle g^mH = eH = H$?

    But if $\displaystyle g^mH = H$ then $\displaystyle g^m$ must $\displaystyle \in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $\displaystyle h \in H$).
    Yes, that is correct. $\displaystyle \left|G/H\right|=[G:H]$ and so $\displaystyle \left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2010
    Posts
    37
    Quote Originally Posted by Drexel28 View Post
    Yes, that is correct. $\displaystyle \left|G/H\right|=[G:H]$ and so $\displaystyle \left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.
    *dances* Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quotient Groups - Infinite Groups, finite orders
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Aug 11th 2010, 07:07 AM
  2. Quotient Groups
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: Nov 17th 2009, 04:42 AM
  3. Quotient groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 28th 2009, 06:49 PM
  4. Quotient groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 24th 2009, 11:58 AM
  5. quotient groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Sep 22nd 2009, 03:19 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum