# Isomorphism from permutation of a group to the group itself.

The question is taken from Section 9, page 101 of 'A First Course in Abstract Algebra' by John B. Fraleigh, 7th edition.

"Let $G$ be a group. Prove that the permutations $\rho_{a} : G \rightarrow G$, where $\rho_{a}(x) = xa$ for $a \in G$ and $x \in G$, do form a group isomorphic to G."

My attempt at an answer: 1) Proving closure $\rho_{a}(x) = xa$, since $x \in G$ and $a \in G$, $xa \in G \Rightarrow$ closure.

2) Proving one-to-one: Let $\rho_{a}(x) = \rho_{a}(y)$ for some $x,y\in G$. Then $xa = ya$. Since $a \in G, \exists a^{-1} \in G$ such that $a.a^{-1} = a^{-1}.a = e$, where $e$ is the identity. Therefore, by multiplying on the right by $a^{-1}$, $x=y$.

Since $\rho_{a}(x) = \rho_{a}(y) \Rightarrow x=y$, this is one-to-one.

3) The last part is where I am running into problems. I think I am trying to prove $\phi: \rho_{a} \rightarrow G$, and $\phi(xy) = \phi(x)\phi(y)$. So... $\rho_{a}(xy) = \rho_{a}(x)\rho_{a}(y)$.

However, this is $xya$ on the left hand side and $xa.ya$ on the right hand side.

Any help would be appreciated.

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Did you try to prove that $\psi:a\mapsto \rho_a$ is an isomorphism. Prove that it's well defined, homomorphism and is a bijection. – Sigur Oct 17 '12 at 0:14
When it says, prove the permutations form a group, it means, a group under the operation of composition. So, for example, for closure, you have to show that if $\rho$ and $\sigma$ are permutations then so is $\rho\circ\sigma$. – Gerry Myerson Oct 17 '12 at 0:16

Let $S$ denote the group of the given permutations: $$S:=\big(\{ \rho_a \mid a\in G \}, \circ\big)$$ As it is said in the comment, the operation of $S$ is not 'multiplication' but function composition, that is: $\rho_a\circ \rho_b (x) = \rho_a(\rho_b(x))$.
And the isomorphism will be $\rho$ itself: $G\to S$, or if it doesn't fit because of the order, then perhaps $a\mapsto \rho_{a^{-1}}$. (Anyway, what is your $\phi$??)
By definition of $S$, $\rho$ is surjective, and so on...
Hi, thanks for your answer. Just to clarify, is the binary operation function composition because each 'function' is a permutation homomorphism for $a$, where $a \in G$? – achacttn Oct 17 '12 at 2:48