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Sandy 發問於 科學及數學數學 · 1 十年前

Math Group

1. Let G be a group with |G| = 6.

(a) Prove: G contains elements of order 2 and 3.

(b) Prove: If G is not cyclic, then G has exactly three elements of order 2: r1, r2, and r3.

(c) Prove: If G is not cyclic, then G and S3 are isomophic group. {Define the map c: G->Sym({r1, r2, r3}) by: c(g)(ri) = g 。ri 。g^-1.

Prove that c is an isomorphism of groups.

2. Let G be a group with |4| = 4. Define the map A: G-> Sym(G) by A(g)(x) = g。x for all g in group G and all x in the set G.

(a) Prove: A defines an isomorphism between G and the subgroup A(G) Sym(G).

(b) Prove Either G is cyclic or G is isomorphics to V4.

1 個解答

評分
  • 1 十年前
    最愛解答

    Q1 first

    1.(a)

    It follows from Sylow’s theorem that there is subgroup of order 2 and order 3. As 2 and 3 are prime numbers, the non-identity element of such subgroups are of order 2 and 3 respectively.

    (b)

    Let a be an element of order 3 and r1 be an element of order 2,

    ar1 =/= r1a

    Otherwise ar1 is an element of order 6 which contradicts the fact that G is not cyclic.

    therefore

    r2 = ar1a^(-1) = ar1a^2 is not equal to r1 and is of order 2.

    Similarly r3 = a^2r1a^(-2) = a^2r1a is of order 2 and not equal to r1 and r2. There are at least 3 distinct elements of order 2.

    As a, a^2 and e are other distinct elements of the group, there cannot be more than 3 distinct elements of order 2.

    Hence, there are exactly 3 elements: r1, r2, and r3.

    (c)

    Define the map c: G->Sym({r1, r2, r3}) by: c(g)(ri) = g 。ri 。g^-1.

    c(g1。g2) (ri)= g1。g2 。ri 。(g1。g2)^-1

    = g1。g2 。ri 。g2^-1。g1^-1

    = g1。c(g2)(ri) 。g1^-1

    =c(g1) 。c(g2)(ri)

    Now if c(g) = id,

    Then g。ri。g^-1 = ri for all ri

    If g is of order3, it is not possible from similar argument as in (b).

    If g is of order 2, WLOG, let g=r1,

    r1r2r1^-1=r2

    Therefore

    r1r2=r2r1

    then r1r2 is of order 2 (r1r2r1r2=r1r2r2r1=r1r1=e)

    r1r2=/=r1 and r1r2=/=r2, then r1r2=r3.

    Then {e, r1, r2, r3} is a subgroup of G, then |G| is divisible by 4. Contracting the fact that |G| is 6.

    c is an isomorphism of groups.

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