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Math Help - orders/index, Prove [G:H]=[G:K][K:H]

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    orders/index, Prove [G:H]=[G:K][K:H]

    Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by apple2009 View Post
    Suppose H and K are both subgroup of a finite group G, and H⊂K. Prove [G:H]=[G:K][K:H]
    In a sense here, we see that H\leq K\leq G. It follows by Lagrange's Theorem that both \left|H\right| and \left|K\right| divide \left|G\right|.

    Therefore,

    \left[G:K\right]=\frac{\left|G\right|}{\left|K\right|}\implies \left|G\right|=\left[G:K\right]\left|K\right|

    \left[G:H\right]=\frac{\left|G\right|}{\left|H\right|}\implies\lef  t|G\right|=\left[G:H\right]\left|H\right|

    \left[K:H\right]=\frac{\left|K\right|}{\left|H\right|}\implies\lef  t|K\right|=\left[K:H\right]\left|H\right|.

    Substituting the third equation into the first, we get

    \left|G\right|=\left[G:K\right]\left[K:H\right]\left|H\right|.

    Now substituting this result into the second equation, we have

    \left[G:K\right]\left[K:H\right]\left|H\right|=\left[G:H\right]\left|H\right|\implies \left[G:H\right]=\left[G:K\right]\left[K:H\right].

    Does this make sense?
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  4. #4
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    a less trivial version of this problem is to let G be any group (not necessarily finite) and assume that [G : H] is finite. note that then both [G : K] and [K : H] will also be finite.
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