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Math Help - Quotient Groups

  1. #1
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    Quotient Groups

    Let H be a normal subgroup of finite group G. If the order of the quotient groups G/H is m, prove that g^m is in H for all 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
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  2. #2
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    Hint: Given any group G of order n and any g \in G, g^n = e.

    In the case of G/H, what does this tell you about the coset g^mH?
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  3. #3
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    Quote Originally Posted by spoon737 View Post
    Hint: Given any group G of order n and any g \in G, g^n = e.

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

    But if g^mH = H then g^m must \in H by theorem (If H is any subgroup of a group G, then hH = H = Hh, h \in H).
    Last edited by MissMousey; May 4th 2010 at 02:40 PM.
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  4. #4
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    Quote Originally Posted by MissMousey View Post
    Since G/H is of finite order m, does that imply that g^mH = eH = H?

    But if g^mH = H then g^m must \in H by theorem (If H is any subgroup of a group G, then hH = H = Hh, h \in H).
    Yes, that is correct. \left|G/H\right|=[G:H] and so \left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}} and what you said finishes it.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Yes, that is correct. \left|G/H\right|=[G:H] and so \left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}} and what you said finishes it.
    *dances* Thank you!
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