In Pinter's A Book of Abstract Algebra, the reader is asked in Chapter 15 Exercise C5 to prove the following:
I've translated this (for my own benefit) into the following notation:
$\exists x \in G \backepsilon \langle Hx \rangle = G / H \iff \exists a \in G \backepsilon \forall g \in G, \exists n \in \mathbb Z \backepsilon ga^n \in H$
where the backwards epsilon is my notation for "s.t." and $H$ is a normal subgroup of $G$. Note that I changed the symbol-notation from $x$ to $g$ to avoid any confusion. I will adhere to the highlighted notation in my below steps.
The only theorem that I am also equipped with is the following:
$Ha=Hb \iff b \in Ha \ \land \ a\in Hb$
Pinter does not mention whether or not the elements have finite or infinite order…so I will show what I have thus far for the case of quotient groups whose elements have finite order (I have no clue how to even begin in the case of quotient groups whose elements have infinite order).
So…without further adieu:
Let $Hx$ be the coset that generates all other cosets in $G/H$.
Let $\operatorname{ord}(Hx)=m \ \ \ \ \ $(this is where the finite part came in)
Therefore $(Hx)^m = He$
Which means $(Hx^m) = H$
… and $x^m \in H$
This is where I get stuck. It seems to me the next step is to show that $\exists a \backepsilon \forall g, \exists n \backepsilon ga^n = x^m$ or $x^{mk}$, where $k$ is an integer. I include the $x^{mk}$ because if $x^m \in H$ then $x^{mk} \in H$ as well.
I still haven't used the fact that $Hx$ generates $G/H$ so I am guessing that $G/H = \{H, Hx, Hx^2, Hx^3, …, Hx^{m-2}, Hx^{m-1} \}$ is probably relevant.
Any thoughts? (Also, if there is a more general case that can be reasoned for infinite order, please feel free to comment). Cheers~
Best Answer
Suppose $G/H $ is cyclic, i.e. $G/H=\langle aH\rangle$ for some $a\in G $. Let $x\in G $. Then $x^{-1}H $ can be written as $a^nH $ for some positive integer $n$.But $x^{-1}H=a^nH$ iff $(x^{-1})^{-1}a^n=xa^n\in H $. Can you complete now?