The commutator subgroup centralizes every cyclic normal subgroup

Suppose K \leq G is a normal subgroup and that K is cyclic. Prove that G^\prime \leq C_G(K), where G^\prime denotes the commutator subgroup. [Hint: Recall that the automorphism group of a cyclic group is abelian.]

We let \sigma_a denote the automorphism “conjugate by a“.

Let x,y \in G and k \in K. Now [x,y]k[x,y]^{-1} = [x,y]k[y,x] = x^{-1}y^{-1}xyky^{-1}x^{-1}yx = (\sigma_x \circ \sigma_y \circ \sigma_{x^{-1}} \circ \sigma_{y^{-1}})(k) = (\sigma_{x} \circ \sigma_{x^{-1}} \circ \sigma_y \circ \sigma_{y^{-1}})(k) = (\sigma_{xx^{-1}} \circ \sigma_{yy^{-1}})(k) = k, since the automorphism group of K is abelian. Thus [x,y]k = k[x,y], and we have G^\prime \leq C_G(K).

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  • Gobi Ree  On January 4, 2012 at 1:24 am

    Note that the hint is from Proposition16 in section4.4, or exercise 2.3#25, which states that \mathsf{Aut}(Z_n) \simeq (\mathbb{Z}/n\mathbb{Z})^\times.

    Also note that since K is normal, \sigma_x is an automorphism of K.

  • kem  On March 15, 2012 at 10:22 am

    I think you have a typo (missing = ) in the 2nd to last line.

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