Dummit and Foote 4.4.13

Let $G$ be a group of order 203. Prove that if $H$ is a normal subgroup of order 7 in $G$ then $H \leq Z(G)$. Deduce that $G$ is Abelian.
In symbols, we may represent this problem as follows:

$\Bigg\lbrace \begin{array}{cc} |G|=203 & & \\ H \unlhd G & \Longrightarrow H \leq Z(G) \Longrightarrow G \quad Abelian \\ |H|=7 \\ \end{array}$

Proof: First note that $203=7 \cdot 29$ a square-free prime factorization. By Corollary 15, we have $N_G(H)/C_G(H) \lesssim Aut(H)$. Since $H \lhd G$, $N_G(H)=G$. With $|H|=7$, $H \cong \mathbb{Z_7}$.
Corollary 16 gives $Aut(H) \cong (\mathbb{Z}/7\mathbb{Z})^{\times} \cong \mathbb{Z}_7^* \cong \mathbb{Z}_6$. So, $G/C_G(H) \lesssim \mathbb{Z}_6 \Longrightarrow |G/C_G(H)|\huge|$ $6$ by Lagrange. So, $\dfrac{203}{C_G(H)} \huge|$ $6$. $\Longrightarrow |C_G(H)|=203 \Longrightarrow H \leq Z(G)$.

We have that $H \leq Z(G) \leq G$. Since $G$ has the square-free prime factorization $7\cdot29$, $|H|=7$ means that $H$ is a maximal (and normal) subgroup of $G$. So, either $H=Z(G)$ or $Z(G)=G$. However, if it is true that $H=Z(G)$, then $G/Z(G) = G/H$ is of order $\frac{203}{7}=29$ implying $G/H$ is cyclic. Problem 3.1.36 implies that $G$ is Abelian $\Longrightarrow Z(G)=G$ contradicting the assumption that $|Z(G)|=|H|=7$. Thus, is must be the case above that $H < Z(G) = G \Longrightarrow G$ is Abelian.
Additionally, by the Fundamental Theorem of Finite Abelian Groups, $G \cong \mathbb{Z}_7 \bigoplus \mathbb{Z}_{29} \cong \mathbb{Z}_{203} $ $\blacksquare$