# Action on a group descends to an action on its factor group

Let $A$ and $B$ be groups and $N\unlhd A$ is a normal subgroup of $A$. Suppose that $B$ acts on $A$; that is, there exists a group homomorphism (not necessarily monomorphism) $$f:B\to\mathrm{Aut}(A)\text{,}$$ where $\mathrm{Aut}(A)$ is the automorphism group of $A$.

My question: Under what condition does $f$ descends to an action on the factor group $A/N$? That is, under what condition does $f$ induces a group homomorphism $$\overline{f}:B\to\mathrm{Aut}(A/N)\text{?}$$

• Is it that $\forall b\in B, (f(b))(N)\subseteq N$? – Zuriel Nov 29 '14 at 7:30
• Yes it is. Look at here, it might be helpful. en.wikipedia.org/wiki/Characteristic_subgroup – Crostul Nov 29 '14 at 8:27
• Thank you @DerekHolt for the conclusion! In the article arxiv.org/pdf/math/0409307.pdf, for the proof of Proposition 3.1 I think the author basically only proved that $f(b)(N)\subseteq N$ (of course, with different notations), am I correct? – Zuriel Nov 29 '14 at 9:11
• Well you do need the stronger condition $f(b)(N) = N$ for all $b \in B$, but in fact that follows from $f(b)(N) \le N$ and $f(b^{-1})(N) \le N$, so you are correct. Sorry about that! – Derek Holt Nov 29 '14 at 9:12
• Thank you @DerekHolt for the clarification! – Zuriel Nov 29 '14 at 9:14