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Thread: Questions Involving Cosets

  1. #1
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    Questions Involving Cosets

    1. Let G = Q8 (quaternion group). Find the right cosets of H in G for:
    a) H = <J>
    b) H = <-I>

    2. Find the right cosets of the subgroup H = <(1,1)> in Z2 x Z4

    3. Let H be a subgroup of a group G and define H congruent on G by letting x congruent to y IFF x^-1y is in H.
    a) Show that H congruent is an equivalence relation on G
    b) Show that the equivalence classes under H congruent are the left cosets of H in G.
    c) Show that for a,b in G, aH = bH iff a^-1b is in H.

    4. Let G = Q8 (quaternion group). Find [G : H] for H = <-I>, H = <K>, and H = <-L>.

    5. a) In G = (Z48, +), find [G : H] for H = <32>.
    (+ is actually addition mod 48, if it was unclear). There are two more parts to this question but if someone could tell me how to do this one, I can follow with the others.
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  2. #2
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    Quote Originally Posted by Janu42 View Post
    1. Let G = Q8 (quaternion group). Find the right cosets of H in G for:
    a) H = <J>
    We have $\displaystyle H = \{1,-1,j,-j\}$ then $\displaystyle iH=\{i,-i,k,-k\}$.
    Thus, these two sets are the cosets.

    2. Find the right cosets of the subgroup H = <(1,1)> in Z2 x Z4
    First $\displaystyle H=\{ (0,0),(1,1),(0,2),(1,3)\}$ thus $\displaystyle (0,1)+H = \{(0,1),(1,2),(0,3),(1,0)\}$.
    These are the cosets.

    3. Let H be a subgroup of a group G and define H congruent on G by letting x congruent to y IFF x^-1y is in H.
    a) Show that H congruent is an equivalence relation on G
    b) Show that the equivalence classes under H congruent are the left cosets of H in G.
    c) Show that for a,b in G, aH = bH iff a^-1b is in H.
    Of course, $\displaystyle x\equiv x$ since $\displaystyle x^{-1}x = e\in H$ thus it is reflexsive. It is symmetric because if $\displaystyle x\equiv y\implies x^{-1}y\in H$ but then $\displaystyle (x^{-1}y)^{-1}\in H\implies y^{-1}x\in H\implies y\equiv x$. Finally, it is transitive because if $\displaystyle x\equiv y\text{ and }y\equiv z$ this means $\displaystyle x^{-1}y,y^{-1}z\in H$ but then $\displaystyle (x^{-1}y)(y^{-1}z) = x^{-1}z\in H\implies x\equiv z$. This completes the proof that $\displaystyle \equiv$ is an equivalence relation.

    Now we wish to prove $\displaystyle [a] = aH$ where $\displaystyle [a] = \{ x\in H | a\equiv x\}$ and $\displaystyle aH = \{ ah|h\in H\}$. Remember to show two sets are equal show that any element in $\displaystyle [a]$ is an element in $\displaystyle aH$ and any element in $\displaystyle aH$ is in $\displaystyle [a]$. If $\displaystyle x\in [a] \implies a\equiv x \implies a^{-1}x\in H$ and so $\displaystyle a^{-1}x = h$ for some $\displaystyle h\in H$, but then this means $\displaystyle x=ah\in aH$. The reserve direction is straightforward too.

    Remember that $\displaystyle aH = [a]\text{ and }bH = [b]$ so for $\displaystyle [a] = [b]$ we require that $\displaystyle a\equiv b$, and so $\displaystyle a^{-1}b\in H$.

    4. Let G = Q8 (quaternion group). Find [G : H] for H = <-I>, H = <K>, and H = <-L>.

    5. a) In G = (Z48, +), find [G : H] for H = <32>.
    (+ is actually addition mod 48, if it was unclear). There are two more parts to this question but if someone could tell me how to do this one, I can follow with the others.
    Here is a hint for these problems. If $\displaystyle G$ is a finite group then $\displaystyle [G:H] = \tfrac{|G|}{|H|}$.

    This is Mine 12,2th Post!!!
    Last edited by ThePerfectHacker; Mar 14th 2009 at 07:52 PM.
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  3. #3
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    Sorry, I don't understand the answer to #1. You put H and then iJ? I'm sorry if I'm missing something here, but I'm confused...
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  4. #4
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    Quote Originally Posted by Janu42 View Post
    Sorry, I don't understand the answer to #1. You put H and then iJ? I'm sorry if I'm missing something here, but I'm confused...
    Sorry, I meant iH not iJ. It is fixed now.
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