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Let $G$ be an abelian group and $H$ a subgroup of $G$. For each $g \in G$, does there always exist an integer $n$ such that $g^{n} \in H$?

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Yes, $0$... But I guess you want $n\geq 1$, right? –  1015 Mar 3 '13 at 20:43
    
If $G$ every element of $G$ has finite order, this is trivially true (take $n$ to be the order of $g$). –  Avi Steiner Mar 3 '13 at 20:44
    
No. Consider $G = \mathbb{C}^\times$ and $H=\{1\}$. Consider any element $g\in \mathbb{C}$ such that $|g|=1$ but $g$ is not a root of unity. Note that such an element must exist for cardinal reasons. –  vgty6h7uij Mar 3 '13 at 20:46
    
@vgty6h7uij What you wrote is correct, but the requirement that $|g|=1$ is superfluous. Any $g\in\mathbb C^\times$ with $|g|\neq1$ also serves as a counterexample. –  Andreas Blass Mar 3 '13 at 22:03
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7 Answers 7

up vote 10 down vote accepted

If you mean: there exists $n\geq 1$ such that...

Take $G=\mathbb{Z}$ and $H=\{0\}$.

If not, $n=0$ will always do.

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I have a question. Please can you look at my last question? Please:) –  B11b Mar 3 '13 at 21:02
    
@B11 Hello. Which question? –  1015 Mar 3 '13 at 21:03
    
Hello:)) @julien the question is here math.stackexchange.com/questions/319476/… –  B11b Mar 3 '13 at 21:05
    
@B11 The answer there explains how to prove that the four functions are $C^\infty$. To be slightly more explicit, you need to show that the $n$th derivative of $T$ for $x>0$ is of the form $(P_n(x)e^{-1/x})/x^{2n}$ where $P_n$ is a polynomial. This shows that these derivatives tend to $0$ when $x$ tends to $0$, so $T$ is indeed $C^\infty$. And the other functions are built on $T$ in such a way that they remain $C^\infty$. I am not sure about the Taylor question. I'll have to think about it. What are your thoughts? –  1015 Mar 3 '13 at 21:19
    
yes I understand first part. This is so explicit. But I have no idea about Taylor question. No one give even a hint :( but i need to solve this until the end of this Monday.(tomorrow) –  B11b Mar 3 '13 at 21:31
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This is evident that if $[G:H]=n<\infty$ then $\forall g\in G, g^n\in H$.

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Actually since $H$ is normal in $G$, the question is equivalent to the factor $G/H$ having finite order.... –  N. S. Mar 3 '13 at 20:50
    
@N.S.: I think, The OP would be better to ask in two parts. Clearly, it is false as many answer here shout that. –  B. S. Mar 3 '13 at 20:53
    
What I am saying is that the problem is true if and only if $[G:H]< \infty$. So your answer is actually a complete characterization ;) –  N. S. Mar 3 '13 at 20:57
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@N.S. Can $G/H$ be both torsion and infinite? –  anon Mar 3 '13 at 21:25
    
@anon Ups, I should drink my coffee before answering :) –  N. S. Mar 3 '13 at 21:42
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No. Let $G=\mathbb{Z}\times\mathbb{Z}$, $H=\{(x,0)\}$. Then $(0,1)^n\notin H$ for any $n>0$

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Hint: For $n\geq 1$: we can find find a counterexample subgroup $H\leq G$, with $G =(\mathbb{Z}, +)$ which is infinite and cyclic: Put $H = \{e\} = \{0\} \leq G$.

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Very nice hint! +1 –  Amzoti May 11 '13 at 0:58
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No. Let $G = \mathbb{Z}^2$. Let $H = <(1,0)>$. What is $(0,1)^n$?

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If $G$ is periodic, then yes: if $n$ is the order of $g$ then $g^n=1\in H$. Otherwise - no. Let $G=H\oplus K$ is torsionless, $x\in K, x\ne 1$. Then for all $n$ $x^n\not\in H$.

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Consider the circle group $\Bbb{T} = \{e^{ix}\mid x\in\Bbb{R}\}$, and consider the subgroup given by $H = \{z\in\Bbb{T}\mid z^n = 1\textrm{ for some }n\in\Bbb{Z}\setminus\{0\}\}$. $e^i\in\Bbb{T}$ can never satisfy $e^{in} = 1$ for any $n\in\Bbb{Z}\setminus\{0\}$.

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