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Suppose that $G$ is an abelian group and that I want to prove that a certain subgroup $K$ of $G$ has limited exponent and that there exists a normal subgroup $N$ of $G$ with limited exponent. Can I suppose without loss of generality that $N=\{1\}$ and to try to prove that $K$ has limited exponent?

Remark:It is not in my hypothesis that KN/N has exponent.

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It is not in my hypothesis that KN/N has exponent. –  User2040 Apr 29 '12 at 1:26
    
Yes, but in assuming that $N=1$, you are passing to the quotient group. Hence, it now suffices to show that the image of $K$, which is $KN/N$, has finite exponent. –  Brett Frankel Apr 29 '12 at 1:48
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This is a bit strange... there is always a normal subgroup with bounded exponent (the identity); and since $G$ is abelian, if $K$ is nontrivial and has bounded exponent then of course $G$ has a nontrivial normal subgroup that has bounded exponent: $K$ itself. I think you are trying to abstract form the situation you actually have, but the question as phrased is rather strange.... –  Arturo Magidin Apr 29 '12 at 3:01
    
New question:Suppose Let be G a group (no abelian) . I want to prove that a certain subgroup K of G has limited exponent. Suppose that there exists a normal subgroup N (possibly different of K) of G with limited exponent. Can I suppose without loss of generality that N={1} to show that K has limited exponent? –  User2040 Apr 29 '12 at 13:23
    
Ops!Remove the word it supposes in begin of the text. –  User2040 Apr 29 '12 at 13:26

2 Answers 2

What Arturo was pointing out in his comment was that every subgroup of an abelian group is normal, so if you can prove that $K$ has bounded exponent, then you've found a non-trivial normal subgroup with bounded exponent. You can just let $N = K$.

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Yes, you can, if I am understanding your question correctly. If $N$ has exponent $n$ and $KN/N$ has exponent $m$, then for any $k\in K$, $k^m\in N$, so $k^{mn}=(k^m)^n=1$

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