Fly 發問時間: 科學數學 · 9 年前

【代數】simple group

Suppose G is a finite simple group and let n be the smallest positive integer such that |G| | n!

Prove that G cannot have a proper subgroup of index less than n

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  • 9 年前
    最佳解答

    Let H be a subgroup of G and X be the set of all left cosets of H

    Let phi:G ---> Sx ,by phi(g) = sai_g ,where sai_g is defined by

    sai_g:X--->X with sai_g(aH) = (ga)H

    Then we have that

    (i) phi is well-define : It's clear

    (ii) phi is a homomorphism : phi(a)phi(b) = sai_a(sai_b) = sai_ab = phi(ab) since for all g belongs to G, sai_a(sai_b(gH)) = sai_a(bgH) = abgH = sai_ab(gH)

    And ker(phi) is normal in G.Since G is simple, ker(phi) = {e} or G

    But ker(phi) not equals to G since G is not trivial,

    hence we have that ker(phi) ={e}

    And phi is bijective since G is finite

    --->|G| = |Im(phi)| = |Sx| = |X|! = (G:H)! < n!

    That's a contradiction of n is the smallest positive integer such that |G| | n!

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