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 Post subject: group theory question
PostPosted: Sat, 13 Feb 2010 20:01:20 UTC 
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"Let G be a nontrivial group with no proper subgroups except the trivial one. Show that G is finite and that the order of G is prime."

Okay, I think that maybe we know that G is finite because if it was infinite it you could always have an infinite subgroup that operations could not leave? I'm not sure how to prove that and I am totally stumped about how to prove that the order of G is prime.

I Know that if the order of G is even there must be an element z such that z^2=e where e is the identity element. Therefore you could have a subgroup with just {e,a}. But this only proves that finite group with no proper subgroups must have odd order, not necessary prime.

Any help would be appreciated.


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PostPosted: Sun, 14 Feb 2010 06:35:39 UTC 
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Zeta wrote:
"Let G be a nontrivial group with no proper subgroups except the trivial one. Show that G is finite and that the order of G is prime."

Okay, I think that maybe we know that G is finite because if it was infinite it you could always have an infinite subgroup that operations could not leave? I'm not sure how to prove that and I am totally stumped about how to prove that the order of G is prime.

I Know that if the order of G is even there must be an element z such that z^2=e where e is the identity element. Therefore you could have a subgroup with just {e,a}. But this only proves that finite group with no proper subgroups must have odd order, not necessary prime.

Any help would be appreciated.


Pick any non-identity element g. If it has infinite order, then \langle g^2\rangle is a nontrivial proper subgroup of G. If g has finite order, look at the subgroup \langle g\rangle and you get \langle g\rangle=\langle g^2\rangle=\dots=\langle g^{\lvert g\rvert-1}\rangle=G, hence \lvert g\rvert must be a prime.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Sun, 14 Feb 2010 06:45:27 UTC 
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Also, there is a finite simple group of order 2.

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PostPosted: Sun, 14 Feb 2010 17:20:27 UTC 
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outermeasure wrote:

Pick any non-identity element g. If it has infinite order, then \langle g^2\rangle is a nontrivial proper subgroup of G. If g has finite order, look at the subgroup \langle g\rangle and you get \langle g\rangle=\langle g^2\rangle=\dots=\langle g^{\lvert g\rvert-1}\rangle=G, hence \lvert g\rvert must be a prime.


Thank you for the speedy reply but I don't yet understand it. Using the set of integers under addition as an example of an infinite group, lets say that g=4. Than:

g^2=8 So than if we do 8*8 we get 16 which leaves the subgroup, also, This group does not contain the inverse of any of these numbers.

I'm sure I'm probably missing a fundamental fact about what you mean that will clear this up but please help.


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PostPosted: Tue, 16 Feb 2010 20:00:57 UTC 
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Consider the group of integers under addition, when g = 4, then \langle g^2 \rangle = \{ \ldots, -16, -8, 0, 8, 16 \ldots \} = \{ (g^2)^n \mid n \in \mathbb{Z} \}


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PostPosted: Tue, 16 Feb 2010 20:40:19 UTC 
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daveyinaz wrote:
Consider the group of integers under addition, when g = 4, then \langle g^2 \rangle = \{ \ldots, -16, -8, 0, 8, 16 \ldots \} = \{ (g^2)^n \mid n \in \mathbb{Z} \}


Thank you! That clears things up.

I had thought that might be it but it's nice to know for sure.


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