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Let $G$ be a group acting on a set $S$, by means of $(g,s)\mapsto s^g$. If $S$ is itself also a group, then it is natural to impose the further condition that $(st)^g=s^gt^g$. This seems to be the standard notion of a group acting on a group, and the action of a group on itself by conjugation is a central example. But, denoting the group action of $G$ on $S$ by $(g,s)\mapsto g\cdot s$, what about the following natural condition instead: $g\cdot (st)=(g\cdot s) t$ ? The canonical action of a group on itself satisfying this condition. Is there a standard name for such an action? If so, a reference will be appreciated. If not, is there any compelling reason why this notion is not worthy ?

Ittay Weiss
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    If $g.(st)=(g.s)t$, $g(t)=g(e)t$. Let $f(g)=g(e)$. Then $f:G->S$ is an homomorphim. So your group acts by left translation on this group $S$. Is this your question ? – Thomas Jan 09 '16 at 10:40
  • There is the related notion of G-module. – lhf Jan 09 '16 at 10:53
  • Thanks @Thomas for the very clear answer. – Ittay Weiss Jan 12 '16 at 05:29
  • In the OP, I believe "the canonical action of a group on itself" means $g * s := gs$ (left multiplication). I initially assumed it was $g * s = gsg^{-1}$ (conjugation), but that satisfies $g * (st) = (g * s)t$ if and only if $g * t = t$. – jskattt797 May 11 '21 at 19:17

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Just to remove this from the unanswered section, we copy Thomas' comment that answered the OP's question:

If $g \cdot (st)=(g\cdot s)t$,$ g(t)=g(e)t$. Let $f(g)=g(e)$. Then $f: G \to S$ is an homomorphim. So your group acts by left translation on this group $S$. Is this your question?

mathematics2x2life
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  • I'm not sure if "left translation" is the right word, since $g * 1_S$ may vary for different $g \in G$. But yes, for a fixed, individual $g \in G$, $g * t = (g * 1_S) t$ for all $t \in S$, so $g$ acts on $S$ as left multiplication by $g * 1_S \in S$. – jskattt797 May 11 '21 at 19:21