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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 pass to the quotient G/N and to assume without loss of generality we assume that N=1 to show that K has limited exponent?

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The fact that the trivial group has bounded exponent is not very interesting, and hence the fact that $G$ has a normal subgroup of bounded exponent tells you nothing about $G$. What exactly are you actually trying to do? –  Chris Eagle Apr 29 '12 at 13:34
    
How is that going to help you at all, if $N=1$? Then passing to the quotient does nothing. –  Tara B Apr 29 '12 at 14:58
    
The standard way of saying it in English is "bounded exponent", not "limited exponent". –  Arturo Magidin Apr 29 '12 at 20:20
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up vote 2 down vote accepted

You keep talking about "assuming without loss of generality that $N=1$". The fact is that the extra hypothesis that $G$ has a normal subgroup of bounded exponent is vacuous. It gives you absolutely no extra information because every group has a normal subgroup of bounded exponent, namely, the trivial subgroup.

It is true that:

If $G$ is a group, $N$ has bounded exponent, and $KN/N$ (the image of $K$ in the quotient $G/N$) has bounded exponent, then $K$ has bounded exponent.

Which seems to be what you are trying to argue (note that this is not the same as "assuming without loss of generality that $N$ is trivial").

To see the statement above is true, note that $KN/N\cong K/(K\cap N)$. So $K/(K\cap N)$ is of bounded exponent, so there exists $n\gt 0$ such that $k(K\cap N)$ is of exponent $n$ for all $k\in K$. Since $N$ is of bounded exponent, there exists $m\gt 0$ such that for all $n\in N$, $n^m = 1$.

Now let $k\in K$. Then $(k(K\cap N))^m = K\cap N$; so $k^m\in K\cap N$. Since $k^m\in N$, then $(k^m)^n = 1$. Therefore, $k^{nm}=1$ for all $k\in K$, and $nm$ does not depend on $k$. So $K$ is of bounded exponent.

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Thank you! I understood your observations. –  User2040 Apr 29 '12 at 23:51
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