## 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 $G^\prime \leq C_G(K)$.

• 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.

• nbloomf  On March 17, 2012 at 2:41 pm

Thanks!