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$Question$

Let $G$ be an abelian group and let $H$ be a subgroup consisting of all elements of $G$ that have finite order .Prove that every non-identity element in $ G/H $ has infinite order

$ Attempt$

Suppose $G/H$ has a non- identity element of finite order .

Let $g_i=n$ which implies ${g_i}^nH = H$ which implies ${g_i}^n\in H$ which implies $g_i\in H$. Since $g_i$ is an arbitrary element, this implies $H= G$, a contradiction.

Is this proof correct ? If not can someone please give a hint or suggestion on how to move forward.

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  • $\begingroup$ Then $g_i^n$ has finite order, being an element of $H$. $\endgroup$ – Lord Shark the Unknown Jul 1 '18 at 9:12
  • $\begingroup$ Can you please see the edited question. $\endgroup$ – blue boy Jul 1 '18 at 9:14
  • $\begingroup$ You don't explain how $g_i^n\in H$ implies $g_i\in H$. $\endgroup$ – Lord Shark the Unknown Jul 1 '18 at 9:16
  • $\begingroup$ Ok. Thanks.... .. $\endgroup$ – blue boy Jul 1 '18 at 9:20
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You need no contradiction, but you're missing a couple of points.

Let $gH$ be an element of finite order in $G/H$. Therefore there exists $n>0$ with $(gH)^n=H$, that is, $g^nH=H$. Hence $g^n\in H$, so there is $m>0$ with $(g^n)^m=1$. As a consequence $g^{nm}=1$, so $g$ has finite order, hence $g\in H$ and $gH=H$.

Therefore the only element of finite order of $G/H$ is the identity $H$.

The points you are missing: $g^n\in H$ doesn't mean by itself that $g\in H$; there is no assumption that $H\ne G$.

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You should add that ${g_i}^n\in H$ means that $({g_i}^n)^m=e$. But this means that ${g_i}^{mn}=e$. Otherwise, it's fine.

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