On group theory 9
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On group theory 9
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 \(\displaystyle{p}\) .
\(\displaystyle{\left(G,+\right)\simeq \left(\mathbb{Z}_{p},+\right)}\) for some prime number \(\displaystyle{p}\) .
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- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: On group theory 9
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)
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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