Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $g^{n} \in H$?

share|cite|improve this question
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
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.

share|cite|improve this answer
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… – 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

This is evident that if $[G:H]=n<\infty$ then $\forall g\in G, g^n\in H$.

share|cite|improve this answer
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. – Babak 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
@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

No. Let $G=\mathbb{Z}\times\mathbb{Z}$, $H=\{(x,0)\}$. Then $(0,1)^n\notin H$ for any $n>0$

share|cite|improve this answer

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 $H = \{e\} = \{0\} \leq G$.

share|cite|improve this answer
Very nice hint! +1 – Amzoti May 11 '13 at 0:58

No. Let $G = \mathbb{Z}^2$. Let $H = <(1,0)>$. What is $(0,1)^n$?

share|cite|improve this answer

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 $n$ $x^n\not\in H$.

share|cite|improve this answer

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 $n\in\Bbb{Z}\setminus\{0\}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.