Prove that $ \pi_g $ is a permutation

Question

Suppose that $ G $ is group and $ g $ is any element from $ G $ ($g \in G $). We define a function $ \pi_g : G \rightarrow G,\ \ \pi_g(x) = g \cdot x$. Prove that $ \pi_g $ is a permutation of the set $ G $.

Could you show me how to prove this theorem? Thank for your help.

Answer 1

We want to prove that $\pi_g$ is a bijection, that is, a one-to-one onto mapping.

To do this requires proving two things: (i) if $\pi_g(s)=\pi_g(t)$ then $s=t$ and (ii) for any $y\in G$, there is an $x\in G$ such that $\pi_g(x)=y$.

Item (i): Suppose $gs=gt$. Multiplying on the left by the inverse of $g$, we get $s=t$. Item (ii): Your turn!

Answer 2

Hint: you have to prove that the function $\pi_g$ is injective ($gx=gy\implies x=y$) and surjective (all elements arise in the form $gx$).

Answer 3

Can you show that $\pi_{g^{-1}}$ is an inverse of $\pi_g$?