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I'm reading a definition of the holomorph of a group $G$, where it is defined as $G_L\operatorname{Aut}(G)$, where $G_L$ is the group of left translations of $G$, that is, the maps of form $a_L(g)=ag$, for $a,g\in G$.

I assume $G_L\operatorname{Aut}(G)$ is notation for all permutations of $G$ of form $a_L\circ \varphi$ for $a_L\in G_L$ and $\varphi$ an automorphism of $G$. Why is $G_L\operatorname{Aut}(G)$ a group though? It's clear it contains $\mathrm{id}$, but I don't see why it's closed under composition, nor why it contains inverses.

I did manage to notice that $\varphi a_L\varphi^{-1}=\varphi(a)_L$, and so $\varphi^{-1}=a^{-1}_L\varphi^{-1}\varphi(a)_L$, in which case $$ (a_L\varphi)^{-1}=\varphi^{-1}a^{-1}_L=a^{-1}_L\varphi^{-1}\varphi(a)_La^{-1}_L $$ which almost seems like it is in $G_L\operatorname{Aut}(G)$. Thanks.

Source: This is a brief part of question 7 on page 63 of Jacobson's Basic Algebra.

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up vote 1 down vote accepted

As you noticed, $\varphi \circ a_L = \varphi(a)_L \circ \varphi$ for any $a \in G$ and automorphism $\varphi$. Therefore $(a_L \circ \varphi)(b_L \circ \psi) = (a \varphi(b))_L \circ \varphi \psi$. As a composition of transformations, this product is associative. The inverse of $a_L \circ \varphi$ is $\varphi^{-1} \circ a^{-1}_L = \varphi^{-1}(a^{-1})_L \circ \varphi^{-1}$.

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Oh, that all follows quite simply! Thanks WimC. – Adelaide Dokras May 28 '12 at 6:09

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