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Let $G$ be a set with an operation $\odot$ and identity element (i.e. there exist $e \in G$ such that $e \odot g = g \odot e = g$ for all $g \in G$). For $g \in G$, define the left translation $Lg: G \rightarrow G, h \mapsto g \odot h$. Suppose $Lg$ is bijective for each $g \in G$. Prove that if the left translations form a group with respect to function composition, then $G$ is a group.

I can't even prove that $\odot$ is associative.

What I have so far:

  • $\odot$ is associative if and only if $Lg_1 \circ Lg_2 = L(g_1 \odot g_2)$ for all $g_1, g_2 \in G$. This is a trivial result, since $(Lg_1 \circ Lg_2)(h) = g_1 \odot (g_2 \odot h)$ and $L(g_1 \odot g_2)(h) = (g_1 \odot g_2) \odot h$ for all $g_1, g_2, h \in G$.
  • Let $g_1, g_2 \in G$. Since the left translations form a group, there exists $g_3 \in G$ such that $Lg_1 \circ Lg_2 = Lg_3$. Since $Lg_1$ is bijective, there exists $g_4$ such that $g_3 = g_1 \odot g_4$. I don't know how to show $g_4 = g_2$.
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1 Answer 1

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For convenience, I denote $Lg$ by $L_g$.

Let $a,b\in G$.
Then $L_aL_b=L_c$ for some $c\in G$.
Hence $$(L_aL_b)(e)=L_c(e)\implies a(be)=ce\implies ab=c.$$ Thus for every $a,b\in G$, $$L_aL_b=L_{ab}$$ By using the information you obtained, this shows that the operation on $G$ is associative.

Let $a\in G$. Then there exists $b\in G$ such that $$L_aL_b=\iota=L_bL_a.$$ So we have $$(L_aL_b)(e)=\iota(e)=(L_bL_a)(e)\implies L_{ab}(e)=e=L_{ba}(e)\implies (ab)e=e=(ba)e\implies ab=e=ba.$$

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  • $\begingroup$ Thank you so much, Alan! $\endgroup$
    – Yerbolat
    Commented Oct 15, 2020 at 9:25

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