# Give an example..

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• Sep 5th 2009, 05:57 PM
ynj
Give an example..
An infinite group has no proper infinite subgroup. Give an example.
• Sep 5th 2009, 06:52 PM
NonCommAlg
Quote:

Originally Posted by ynj
An infinite group has no proper infinite subgroup. Give an example.

our very own Prufer group: fix a prime number $\displaystyle p$ and define $\displaystyle G=\left \{\frac{m}{p^n} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset \mathbb{Q}/\mathbb{Z}.$ every proper subgroup of $\displaystyle G$ is finite and cyclic. in fact $\displaystyle H_n=<1/p^n + \mathbb{Z}>,$ where the

integer $\displaystyle n \geq 0,$ is fixed, are exactly all proper subgroups of $\displaystyle G.$ this is easy to prove and i'll leave it for you. a standard notation for $\displaystyle G$ is $\displaystyle \mathbb{Z}(p^{\infty}).$
• Sep 5th 2009, 08:33 PM
ynj
Quote:

Originally Posted by NonCommAlg
our very own Prufer group: fix a prime number $\displaystyle p$ and define $\displaystyle G=\left \{\frac{m}{p^n} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset \mathbb{Q}/\mathbb{Z}.$ every proper subgroup of $\displaystyle G$ is finite and cyclic. in fact $\displaystyle H_n=<1/p^n + \mathbb{Z}>,$ where the

integer $\displaystyle n \geq 0,$ is fixed, are exactly all proper subgroups of $\displaystyle G.$ this is easy to prove and i'll leave it for you. a standard notation for $\displaystyle G$ is $\displaystyle \mathbb{Z}(p^{\infty}).$

How about $\displaystyle H=\left \{\frac{m}{p^{2n}} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset G?$
• Sep 5th 2009, 09:10 PM
NonCommAlg
Quote:

Originally Posted by ynj
How about $\displaystyle H=\left \{\frac{m}{p^{2n}} + \mathbb{Z}: \ \ m, n \in \mathbb{Z}, \ n \geq 0 \right \} \subset G?$

this is not a "proper" subgroup because then $\displaystyle \frac{m}{p^{2n-1}} + \mathbb{Z} =\frac{mp}{p^{2n}} + \mathbb{Z} \in H,$ for all integers $\displaystyle m,n$ with $\displaystyle n \geq 1,$ and so $\displaystyle H=G.$
• Sep 5th 2009, 09:32 PM
ynj
Quote:

Originally Posted by NonCommAlg
this is not a "proper" subgroup because then $\displaystyle \frac{m}{p^{2n-1}} + \mathbb{Z} =\frac{mp}{p^{2n}} + \mathbb{Z} \in H,$ for all integers $\displaystyle m,n$ with $\displaystyle n \geq 1,$ and so $\displaystyle H=G.$

hmm..I have got it... If $\displaystyle \frac{p^km}{p^n}$appears, where $\displaystyle \gcd (m,p)=1$, then $\displaystyle \frac{1}{p^{k}},k\leq n-m$appears.. If there is infinite elements in a subgroup, then $\displaystyle n-m\leq M$for every element. And this will be a contradiction...
Great! I have thought of your example before, but I was stuck on my example..Now I get it !