Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)
Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?
Best Answer
No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $\{1\}$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n \neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G \to G'$ such that $f(a) = 1 \iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.