TITLE: Extending Automorphism of $G$ to $G/N$
QUESTION [5 upvotes]: In Herstein's Topics in Algebra, there is a problem:
If $G$ is a group, $T$ an automorphism of $G$, and $N$ a normal subgroup of $G$, s.t. $(N)T \subset N$, construct an automorphism of $G/N$.
My first thought was to simply define $T^{\prime}$: $(Nx)T^{\prime}=N(xT)$, but unfortunately this is not necessarily one to one (e.g. $G=\mathbb{R}$, $N=\mathbb{Z}$, $xT=2x$).
So my question is: Is there some meaningful way to extend $T$ that I should be looking for, or should I simply impose the requirement that $(N)T=N$? If yes, please only give a hint of the direction I should be looking at most, I'm trying to solve these problems myself.
Thanks in advance.
P.S. If the notation's a little weird, that's because I'm using Herstein's notation, in particular the convention of writing mappings on the right.

REPLY [8 votes]: A slightly smaller example should justify your assumption that $(N)T=N$.
Take $G=\mathbb{Q}$, $N=\{ \tfrac{a}{2b+1} : a,b \in \mathbb{Z} \}$, and $T:G\to G:x\mapsto 2x$. Then $G/N$ is a 2-group, so if $x \in G/N$ is such that $(x)T' = 2x$, then $x=N$ is the identity. So $T'$ cannot be defined in any direct fashion from $T$; it will only agree on the identity.
The automorphism group of $G$ is isomorphic to $\mathbb{Q}$ while the automorphism group $G/N$ is isomorphic to the additive group $J_2$ of 2-adic integers. The only homomorphism from $\mathbb{Q}$ to $J_2$ is the zero homomorphism.
In other words, if you require the map from $T$ to $T'$ to be a homomorphism (so that $T'S' = (TS)'$), then $(x)T' = x$ for all $x \in G/N$ and $T \in \operatorname{Aut}(G)$. @xaviermo2's suggestion may be the only sane possibility.