On group theory 9

Groups, Rings, Domains, Modules, etc, Galois theory
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Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

On group theory 9

#1

Post by Papapetros Vaggelis »

Prove that if \(\displaystyle{\left(G,+\right)}\) is an abelian simple group, then

\(\displaystyle{\left(G,+\right)\simeq \left(\mathbb{Z}_{p},+\right)}\) for some prime number \(\displaystyle{p}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: On group theory 9

#2

Post by Papapetros Vaggelis »

We give a solution :

Suppose that \(\displaystyle{\left(G,+\right)}\) is an abelian simple group.

Let \(\displaystyle{x\in G-\left\{0\right\}}\). Then, \(\displaystyle{\langle{x\rangle}=\left\{n\,x\in G: n\in\mathbb{Z}\right\}}\)

is a normal subgroup of \(\displaystyle{\left(G,+\right)}\). Since the group \(\displaystyle{\left(G,+\right)}\)

is simple, we have that \(\displaystyle{\langle{x\rangle}=\left\{0\right\}}\) or \(\displaystyle{\langle{x\rangle}=G}\) .

But, \(\displaystyle{x\neq 0}\), so : \(\displaystyle{G=\langle{x\rangle}}\), that is \(\displaystyle{\left(G,+\right)}\) is cyclic.

If \(\displaystyle{|G|=\infty}\), then, \(\displaystyle{\left(G,+\right)\simeq \left(\mathbb{Z},+\right)}\)

and \(\displaystyle{\left(\mathbb{Z},+\right)}\) is not simple, a contradiction.

Therefore, \(\displaystyle{\left(G,+\right)}\) is a finite, cyclic and simple group, so :

\(\displaystyle{\left(G,+\right)\simeq \left(\mathbb{Z}_{p},+\right)}\) for some prime number \(\displaystyle{p\in\mathbb{N}}\) .

Source : Dummit-Foote (Unaswered exercise)
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