202101141538 Homework 1 (Q4)

$\phi :G_1\rightarrow G_2$ is said to be a group embedding if it is an injective group homomorphism.

For a group $G$ and $a\in G$ , we define the left translation of $G$ by $a$

$l_a:G\rightarrow G$ , $g\mapsto ag,\qquad g\in G$ .

Thus $l$ is a mapping from a value into a function of the value.

I would like to first check that $l$ is a homomorphism,

i.e., $l(a_1a_2)=l(a_1)l(a_2)$ .

$\begin{aligned} \textrm{LHS} & = l(a_1a_2)\\ & =l_{a_1a_2}(g)\qquad\qquad (g\in G)\\ & = a_1a_2g\qquad\qquad (a_1,a_2\in G)\\ & = l_{a_1}(l_{a_2}(g))\\ & =(l_{a_1}\circ l_{a_2})(g)\qquad\qquad (\circ\,\textrm{as the operation in }\textrm{Perm}(G))\\ & = (l_{a_1}l_{a_2})(g)\\ & = l(a_1)l(a_2)\\ & = \textrm{RHS}\\ \end{aligned}$

$\therefore$ $l$ is a homomorphism.

Secondly, I would like to check that $l$ is injective.

If $l(a_1)\neq l(a_2)$ , or $l_{a_1}(g)\neq l_{a_2}(g)$ , or, $a_1g\neq a_2g$ for $a_1,a_2, g\in G$ , then $a_1gg^{-1}\neq a_2gg^{-1}$ guaranteed that the inverse of $g$ always exists in $G$ . It follows that $a_1\neq a_2$ .

$\therefore$ $l$ is injective.

In conclusion, $l$ is a group embedding.

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