# Let $(G,*)$ be a group and $g$ be a fixed element of $G$. Prove that $G=\{g*x \mid x \in G\}$

Let $(G,*)$ be a group and $g$ be a fixed element of $G$. Prove that $G=\{g*x \mid x \in G\}$

My proof:

Let $(G,*)$ be a group and $g$ be a fixed element of $G$.
Let $x=g^{-1}*y$, where $y$ is an arbitrary element of $G$.
Now we can write $g*x=g*(g^{-1}*y)=(g*g^{-1})*y$ by the associative property of groups.
But $(g*g^{-1})*y=e*y=y$, where $e$ is the identity element in $G$.
Therefore any element $y$ of $G$ can be written as $g*x$ and hence $G=\{g*x \mid x \in G\}$

I can't seem to find any holes in the argument but it seems too simple. Any comments? Is this correct?

• Well, you have proved that $G \subseteq \{g*x \mid x \in G\}$. For the sake of completeness, you can mention that $\{g*x \mid x \in G\} \subseteq G$, because of closure (as you have gone to the extent of explicitly demonstrating associativity). – M. Vinay Jul 4 '14 at 5:22
• Thanks! That definitely clarifies things – leibnewtz Jul 4 '14 at 5:26
• You should never, ever need to mention closure explicitly. By definition $\ast$ is a function $G \times G \to G$. – Qiaochu Yuan Jul 4 '14 at 5:36
• @QiaochuYuan Well, a hidden mentioning of closure e.g. in the form of "Trivially, $\{\,g*x\mid x\in G\,\}\subseteq G$" may be justified though. On th eother hand, I even assume that th OP left that point out precisely for that reason: It's trivially true. – Hagen von Eitzen Jul 4 '14 at 6:12
• Don't we just show that the map is a bijection onto $G$? injection: $gx=gy$, then pre-multiply by $g^{-1}$, and onto: for any $w \in G$ ;let $x:=g^{-1}w \rightarrow gx=gg^{-1}w=w$? – user99680 Jul 4 '14 at 6:21

Maybe to summarize all the comments and have an answer to the question, we can just show that the map $$g \rightarrow gx$$ , is a bijection onto $G$:
1) Injection: assume $gx =gy$. Then we premultiply by $g^{-1}$ to conclude $x=y$
2) Onto: let $w \in G$ . Then $g(g^{-1}w)=(gg^{-1})w =w$.
So the map $x \rightarrow gx$ is a bijection onto $G$, so that $G$={$gx: x \in G$} (though showing the map is onto is enough).