Results 1 to 2 of 2

Math Help - proof by contradiction

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    87

    proof by contradiction

    First problem:
    Let A be a group with a normal subgroup of finite index N.
    Suppose h conjugate to k such that $h=a^{-1}ka$ for some a \in A.
    Let $A \rightarrow \bar{A}=A/N$.
    Show that for any $x \in A \backslash \langle h \rangle$, $x \notin \langle h \rangle N$.

    I tried to show by contradiction.
    Suppose that $x \in \langle h \rangle N$.
    It is clear that $z=axa^{-1} \neq k$ since $x \notin \langle h \rangle$.
    Then under the mapping, $\bar{z}=zN=\bar{a}\bar{x}\bar{a}^{-1}$.
    Since $x \in \langle h \rangle N$, we can find some m such that $\bar{x}=\bar{h^{m}}$.
    Well, if a is trivial, we have $\bar{z}=\bar{x}=\bar{h^{m}}=\bar{k^{m}}$.
    If $a \notin N$, $\bar{z}=zN=axa^{-1}N=aNxNa^{-1}N=\bar{a}\bar{h^{m}}\bar{a^{-1}}=\bar{k^{m}}$.
    How about when $a \in N$?
    Is it true that $\bar{z}=\bar{h^{m}} \neq \bar{k^{m}}$?


    Second problem:
    Let A be a group with a normal subgroup of finite index N and $x \in A \backslash \langle h \rangle$ such that $x \in \langle h \rangle N$. Is it true that $x^{-1}h^{-1}xh \notin \langle h \rangle$?

    I tried to show by contradiction. But so far I can't get any contradiction.
    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 deniselim17 View Post
    First problem:
    Let A be a group with a normal subgroup of finite index N.
    Suppose h conjugate to k such that $h=a^{-1}ka$ for some a \in A.
    Let $A \rightarrow \bar{A}=A/N$.
    Show that for any $x \in A \backslash \langle h \rangle$, $x \notin \langle h \rangle N$.
    I am a tad confused with what you are trying to show here. I cannot see where the k fits in...

    I interpreted it as: if x\in A\<h> then x \not\equiv h^n mod N for all n (x lies in a different coset of A/N form any power of h).

    However, this is not true. Take A=C_2xC_2, and let N=<(2, 2)>. Take H=<(2, 0)> and let x=(0, 2). Then (0, 2)=(2, 0) mod N...so I think I have interpreted this incorrectly...but I can't see any other way of interpreting it...

    For your second question, and abelian group will do (as then x^{-1}h^{-1}xh=1). The example I gave above works here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof by contradiction
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: February 28th 2011, 10:06 AM
  2. Proof by Contradiction
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: September 28th 2010, 09:23 PM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 27th 2010, 10:07 PM
  4. Proof by contradiction
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: February 5th 2010, 04:17 PM
  5. Proof by contradiction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: April 11th 2009, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum