Relating properties of $H$ to properties of $G/H$

Question

$G$ is a group and $H$ is a normal subgroup of $G$. Prove that $G/H$ is cyclic iff there is an element $a \in G$ with the following property: for every $x \in G$, there is some integer $n$ such that $xa^n \in H$.

Can anyone help, since I have no idea how to work on this question! Thanks sooooo much!

What is $a$? An element of $G$? – Alex Wertheim Jun 05 '13 at 04:39 — Alex Wertheim, Jun 05 '13 at 04:39
yes a belong to G, it is an element belong to G – megan Jun 05 '13 at 04:40 — megan, Jun 05 '13 at 04:40

score 7 · Accepted Answer · answered Jun 05 '13 at 05:03

7

Since $H$ is a normal subgroup of $G$ so $G/H$ defines a group itself. Let $G/H$ be cyclic group, so there is an element $gH\in G/H, ~~g\in G$ that $$G/H=\langle gH\rangle$$ What does this latter thing mean? It means that if $g'\in G$ and so $g'H\in G/H$, we can have: $$g'H=(gH)^k$$ for some integers $k$. Equivalently, we have $$g'H=g^kH$$ (Why?). Because $H$ is normal in $G$. Well, so we have: $$g'(g^k)^{-1}H=H$$ so we have $$g'(g^k)^{-1}\in H$$ for some integers $k$. Conversely,...

answered Jun 05 '13 at 05:03

Mikasa

67,374

there are some integers k: g'g^k belong to H – megan Jun 05 '13 at 05:08
Nice work, my friend! +1 – amWhy Jun 05 '13 at 05:08
you explain super clear thanks so a lot – megan Jun 05 '13 at 05:08
@megan: I am glad, I could help you. :-) Thanks Amy, mu dear friend. – Mikasa Jun 05 '13 at 05:17
@BabakS.: You are doing very nice explanations! +1 – Amzoti Jun 05 '13 at 12:26
@Amzoti: Thanks my dear friend. – Mikasa Jun 05 '13 at 13:12
@Babak S. : In the question, it said to prove that $xa^{n} \in H$. However you proven that $g' (g^{k})^{-1} \in H$. How do you know this proof is complete? – Nov 27 '16 at 21:45

Relating properties of $H$ to properties of $G/H$

1 Answers1