Let $G/H$ be any quotient group and $a^r, a^s$ elements of $G$. If $x \in a^rH \cap a^sH,$ then is $r = s$?

Question

I have the following case: Let $G/H = \langle aH \rangle$ be any quotient group and $a^r, a^s$ elements of $G$. If $x \in a^rH \cap a^sH,$ then is $r = s$?

Well, I thought of this: if $x \in a^rH \cap a^sH,$ then $a^rH \cap a^sH \neq \emptyset.$ So $a^rH = a^sH$. I stopped here and I'm not getting such equality.

If $G/H$ is cyclic, do we have the result?

Answer 1

Let $xH\in G/H$ have order $n$. Then

$$\begin{align} x^{n+m}H&=(xH)^{n+m}\\ &=(xH)^n(xH)^m\\ &=e_{G/H}(xH)^m\\ &=(xH)^m\\ &=x^mH \end{align}$$

for all $m\in\Bbb Z$.

Answer 2

Here the group being a quotient is not essential for your question. Basically you are asking: let $G$ be a group, and $g\in G$, $m,n\in\Bbb Z$ such that $g^m=g^n$. Does it implies $m=n$? Well, if $g=e$ certainly not, as then $g^m=g^n$ holds for every pair $(m,n)$. Let's then suppose $g\ne e$. Now, $g^m=g^n \iff$ $g^{m-n}=e$. If $g$ has finite order $o(g)$, then this just implies $o(g)\mid m-n$, not else (let alone $m=n$). If $g$ has infinite order, then the implication holds true, as by definition of infinite order there isn't any (least) positive integer $k$ such that $g^k=e$. So your implication holds true if $G\cong\Bbb Z$. Now think of $G$ here as your "$G/H$": the claim holds true if $G/H\cong\Bbb Z$ and $a\notin H$ (this latter to mimick my "$g\ne e$").