Homomorphism between a group and its symmentric group.

[bloodhound]

I'm having trouble proving the following:

 

Let [math]G[/math] be a group and let [math]a\in G[/math]. Then

 

[math]\lambda_{a} \colon G \to G[/math]

[math]g \mapsto ag[/math] defines a bijection

 

For a group [math]G[/math], the family of bijections [math]\{\lambda_{g} \colon g \in G\}[/math].

 

The map

 

[math]\lambda \colon G \to SymG[/math]

[math]g \mapsto \lambda_{g}[/math]

is an injective group homomorphism

 

i can do the lambda being bijective part. but i can't show the map from G to Sym G defined as above is an injective homomorphism.

[stevem]

i can't show the map from G to Sym G defined as above is an injective homomorphism.

Not sure which you can't prove - that [math]\lambda[/math] is an injection or that it is a homomorphism or both so here is how to start for both these steps:

 

1. Let [math]a,b \in G[/math] and [math]\lambda_a=\lambda_b[/math]. Apply these 2 functions [math]\lambda_a,\lambda_b[/math] to the identity of G and what can you deduce?

 

2. The binary operation in Sym(G) is composition of functions so for any g in G

[math]\left(\lambda_a \circ \lambda_b\right)(g)=\lambda_a\left(\lambda_b(g) \right)=\lambda_a\left(bg\right) = \dots[/math]

Then use the fact that 2 functions [math]\alpha,\beta[/math] in Sym(G) are equal if and only if [math]\alpha(g)=\beta(g)[/math] for all [math]g \in G[/math]

 

Finally, apply what you have shown to give results about [math]\lambda[/math] using [math]\lambda(g)=\lambda_g[/math]

 

Hope that helps.