Is it always $|G:gG|=1$ where $G$ is a group and $g \in G$?

Question

Is it always $|G:gG|=1$ where $G$ is a group and $g \in G$?

When $G$ is finite then I use Lagrange’s Theorem,

$$|G:gG|=\frac{|G|}{|gG|}=1.$$

When $G$ is not finite then I don't know what should I do. At first I thought about isomorphism $G\simeq gG$ but I don't know if it will help in any way.

Answer 1

Yes.

Lemma: Let $G$ be a group with $g\in G$. Then $G=gG$.

Proof: Let $h\in gG$. Then $h=gk$, some $k\in G$. But then $h\in G$ as $G$ is closed. Hence $gG\subseteq G$.

Let $h\in G.$ Then $h=g(g^{-1}h)\in gG$ because $g^{-1}h\in G$. Thus $G\subseteq gG$.

Therefore, $G=gG$. $\square$

Lemma: Let $G$ be a group. Then $[G:G]=1$.

Proof: By definition, $[G:G]=|G/G|$. But $G/G$ is trivial. $\square$

Theorem: Given a group $G$ and $g\in G$, $$[G:gG]=1.$$

Proof: Since $G=gG$, we have $[G:gG]=[G:G]=1$. $\square$