I have been reading about Cayley's Theorem in the second edition of Artin's Abstract Algebra book. I also read this post Can someone explain Cayley's Theorem step by step?. Which explains the theorem in an intuitive way. I want to understand why exactly is the action of left multiplication an injective homomorphism. In the book they say that it follows from the fact that the operation is faithful. But why a faithful operation is always a injective homomorphism?
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That's an odd comment; the reason is actually associativity.
It's a morphism for the following reason : let $\lambda_g : x\mapsto gx$.
Then $\lambda_g (\lambda_h (x)) = \lambda_g(hx) = g(hx) = (gh)x = \lambda_{gh}(x)$ and therefore, $\lambda_g\circ \lambda_h = \lambda_{gh}$. But that's precisely the definition of a morphism $G\to \mathfrak{S}(G)$ !
Maxime Ramzi
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I think the question is why $g \mapsto \lambda_g$ is injective. – manthanomen Nov 08 '17 at 07:40
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@manthanomen : well there was an ambiguity in the question, and the title only mentioned "homomorphism" so I decided to focus on that. If the OP wants more information about injectivity, he/she can ask me, or see the other answer – Maxime Ramzi Nov 08 '17 at 17:26
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As the other answer states, the fact that $\lambda_g$ is a homomorphism does not require faithfulness. It is injectivity that follows from the fact that the action is faithful. Given distinct $g,h \in G$ (i.e., $g \neq h$) then there exists $x \in G$ such that $\lambda_g(x) = gx \neq hx = \lambda(x)$, which shows that $\lambda_g \neq \lambda_h$.
But mentioning faithfulness at all is a bit strange since we can just take $x = 1$ in the above: given $g,h \in G$ such that $\lambda_g = \lambda_h$, then in particular $$ g = g\cdot 1 = \lambda_g(1) = \lambda_h(1) = h \cdot 1 = h $$ which seems like an even clearer way to prove injectivity.
Viktor Vaughn
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