Exercise 8.4.1

Proof. Let $G$ be a non trivial finite Abelian group.

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If $G\simeq \mathbb{Z}∕\mathit{p\mathbb{Z}}$, $G$ is cyclic of order $p$. Every subgroup $H$ of $G$ has a cardinality dividing $p$, so its order is 1 or $p$, therefore $H=\{e\}$ or $H=G$. The only subgroups of $G$, normal or not, are $\{e\}$ or $G$. So $G$ is simple.

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Suppose that $G$ is a non trivial finite Abelian simple group. As $G$ is Abelian, every subgroup of $G$ is normal in $G$, thus $G$ has no other subgroup that $\{e\}$ or $G$. As $G$ is not trivial, there exists $x\in G,x\ne e$. Then $⟨x⟩$ is a subgroup of $G$ with cardinality greater than 1, therefore $G=⟨x⟩$. $G$ being cyclic, it is isomorphic to $\mathbb{Z}∕\mathit{n\mathbb{Z}},n\in \mathbb{N},n>1$. If $n$ was not prime, $n$ would be divisible by some integer $d,1<d<n$. Then $⟨{[d]}_n⟩$ is a subgroup of $\mathbb{Z}∕\mathit{n\mathbb{Z}}$ of order $n∕d,1<n∕d<n$, so $\mathbb{Z}∕\mathit{n\mathbb{Z}}$ would have a non trivial subgroup, and also $G$. This is a contradiction, so $n=p$ is prime:

$G\simeq \mathbb{Z}∕\mathit{p\mathbb{Z}}$, $p$ prime.