Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Group Theory Proof -- Simple?

Suppose that a and b are elements of a group G with a having odd order. Assuming that aba^-1 -= b^-1 prove that b^2 = e.

Update:

should be aba^-1 = b^-1 not aba^-1 -= b^-1

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  • Duke
    Lv 7
    1 decade ago
    Favorite Answer

    You know of course that (xyz)⁻¹ = z⁻¹ y⁻¹ x⁻¹ for any x, y, z of G, so inverting

    aba⁻¹ = b⁻¹ we get (aba⁻¹)⁻¹ = (b⁻¹)⁻¹ or

    (1) ab⁻¹a⁻¹ = b.

    Let a's order is 2n+1 /odd number for n = 0, 1, 2, , /, that means

    a²ⁿ⁺¹ = e, but then, of course

    (a⁻¹)²ⁿ⁺¹ = (a²ⁿ⁺¹)⁻¹ = a⁻²ⁿ⁻¹ = e also.

    Multiply aba⁻¹ = b⁻¹ by a from left, by a⁻¹ from right:

    aaba⁻¹a⁻¹ = ab⁻¹a⁻¹, but the right side here according (1) is equal to b, i.e:

    a² b a⁻² = b, repeat once again and apply the identity:

    a³ b a⁻³ = aba⁻¹ = b⁻¹, go on the same way:

    a⁴ b a⁻⁴ = b;

    a⁵ b a⁻⁵ = b⁻¹;

    . . . . . . . . . and, after ODD NUMBER OF TIMES

    a²ⁿ⁺¹ b a⁻²ⁿ⁻¹ = b⁻¹

    But a²ⁿ⁺¹ = a⁻²ⁿ⁻¹ = e, so on the above text line we have b = b⁻¹.and then

    b² = bb⁻¹ = e as required.

    P.S. dCromley, it is correct above: a^ord(a) = a^|a| = e

  • 1 decade ago

    I sent an email to Duke asking:

    isn't a²ⁿ⁺¹ = a not =e?

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