- Thread starter
- #1

- Apr 14, 2013

- 4,028

Let $G$ be a group and let $g\in G$. We define: \begin{align*}&\lambda_g:G\rightarrow G, \ x\mapsto gx \\ &\gamma_g:G\rightarrow G, \ x\mapsto gxg^{-1}\end{align*}

Show for all $g,h\in G$:

- $\lambda_{1_G}=\text{id}_G$ and $\lambda_{gh}=\lambda_g\circ \lambda_h$
- $\lambda_g$ is a permutation and it holds that $\lambda_g^{-1}=\lambda_{g^{-1}}$
- The map $\lambda:G\rightarrow \text{Sym}(G), \ g\mapsto \lambda_g$ is a group monomorphism
- $\gamma_{1_G}=\text{id}_G$ and $\gamma_{gh}=\gamma_g\circ \gamma_h$
- $\gamma_g\in \text{Sym}(G)$ and it holds that $\gamma_g^{-1}=\gamma_{g^{-1}}$
- For all $x\in G$ it holds that $\gamma_x(gh)=\gamma_x(g)\gamma_x(h)$
- The map $\gamma:G\rightarrow \text{Sym}(G), \ g\mapsto \gamma_g$ is a group homomorphism
- Determine $\ker (\gamma)$

First of all, to understand the operation... At the map $\lambda_g:G\rightarrow G, \ x\mapsto gx$ the $gx$ is a multiplication of functions or is it a composition or is it like $g(x)$ ?

Because at the first question we have $g=1_G$. Then we have for some $x\in G$ that $\lambda_{1_G}(x)=1_Gx$. What operation is this?