Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ and $H$ be locally compact totally disconnected abelian groups, and $f:G\rightarrow H$ a surjective open map. Let $Y\subseteq G$ be a discrete subgroup in the subspace topology. Is it true that the image $f(Y)$ is also discrete in the subspace topology? If so, how can one prove it?

share|improve this question
    
Thanks for your quick replies. But I forgot one more assumption. The space $H$ is non-compact. Also as is pointed out in the previous answer, the map $f$ is a homomorphism. Does anyone know the answer? –  user32758 Apr 2 '13 at 14:22
    
Can´t you just add an extra $\mathbb{Z}$ factor to $G$ and $H$ and then let $Y$ be generated by the element $(1,0,1,1,1,\dots)$? –  Ramiro de la Vega Apr 2 '13 at 14:37
    
It seems right. Thanks. –  Windi Apr 2 '13 at 16:25

1 Answer 1

No, that's false. You didn't say that $f$ is a homomorphism, but the answer is still no if we require this.

Let $G = {\bf Z} \times C_2 \times C_3 \times C_5 \times \cdots$ be the product of the infinite cyclic group and the cyclic groups of all prime orders. Let $H = C_2 \times C_3 \times C_5 \times \cdots$ and let $f: G \to H$ be the obvious homomorphism with kernel ${\bf Z}$. Let $Y$ be the subgroup of $G$ generated by the element $(1, 1, 1, \ldots)$. It is a discrete infinite cyclic subgroup, but its image in $H$ is not discrete (in fact, it is dense in $H$).

share|improve this answer
1  
Here is a simpler one: if $t\in\mathbb{R}$ is irrational, the image of $t\mathbb{Z}$ under the projection $\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ is not discrete. –  Laurent Moret-Bailly Apr 2 '13 at 7:52
1  
@Laurent: The question asked for $G$ and $H$ to be totally disconnected. –  Nik Weaver Apr 2 '13 at 8:20
    
@Nik: Oops, sorry. –  Laurent Moret-Bailly Apr 2 '13 at 9:26

Your Answer

 
discard

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.