Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Problem involving centralizer of the group?

Suppose A belongs to a group and |A|=5. Prove that C(A) = C(A^3). Find an element a from some group such that |A|=6 and C(A) doesnt equal C(A^3)

I understand C(A) = { gεG : gA = Ag}.....so g and A commute, but I havent seen a good example of a problem like this

I know A^5 = e...so does saying C(A) = C(A^3) imply that a^15=e? I would guess so...or at least something along those lines. also a general question....if set/group has a finite integer order, does that mean it is cyclic?

Thanks a lot

Update:

How does that relate to part b?

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  • 1 decade ago
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    It's easy. If g commutes with A, then g commutes with A^3, because gAAA = AgAA = AAgA = AAAg.

    If g commutes with A^3, then it commutes with A^6 = A^3 A^3, but A^6 = A, so it commutes with A.

    Therefore, g commutes with A^3 iff it commutes with A, so C(A) = C(A^3).

    However, the argument does not work when three divides the order of the element.

    Take, for example, rotation by an angle that generates all other rotations as A, in D_6, the dihedral group of order 12; in other words, the symmetries of a regular hexagon. Then A^3 commutes with every element of the group, but A does not commute with reflections.

    If a group has a finite integer order, it is not necessarily cyclic. The quaternion group and the dihedral groups are not cyclic; nor are the symmetry groups.

    If a group is of prime order, though, then it *is* cyclic.

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