Defining an automorphism of $G/N$ using the automorphism defined on $G$.

Question

This is a problem from Herstein and I do know that there is a similar question uploaded already, but I felt something strange with some assumptions of the question: "Let $G$ be a group, and $T$ be an automorphism of $G$. If $N$ is a normal subgroup of $G$ such that $T(N) \subset N$, then show how you could use $T$ to define an automorphism of $G/N$." I have just tried the most natural mapping $\phi : G/N \to G/N$ defined by $\phi(gN) = T(g)N$, and succeeded in showing that $\phi$ is in fact a well-defined onto homomorphism. What is left is, to show that, $\phi$ is injective. I felt that if $T(N) =N$ was the case, then $\phi$ is clearly an automorphism. But in general, $T(N) \neq N$ unless $G$ is finite(or with more lenient case, $[G:N] <\infty$. If $[G:N]$ was finite, I could have just make use of pigeonhole principle so that $\phi$ is the desired mapping.) I wonder if there is any other method to show that $\phi$ is injective, or we could have defined another new mapping, or the question has failed to add the finiteness of $G$(or $[G:N]$). Any helps or comments will be appreciated. Thanks!

Answer 1

While this is not a solution to the given exercise, I'll give an example to show that your proposed definition of $\phi$ does not always yield an injective function.

Fix a prime $p$.

Let $G,N,T$ be defined as follows: $$ \left\lbrace \begin{align*} &G=(\mathbb{Q},+)\\[4pt] &N=\left\{{\small{\frac{a}{b}}}\in\mathbb{Q}{\;\,\colon\;}p{\,\not\mid\,}b\right\}\\[4pt] &T(x)=px\\[4pt] \end{align*} \right. $$ Then $T$ is an automorphism of $G$ and $T(N)\subset N$.

But if $\phi:G/N \to G/N$ is defined by $\phi(g+N)=T(g)+N$, then letting $g={\large{\frac{1}{p}}}$, we get

• $g\not\in N$ so $g+N\ne N$, hence $g+N\ne 0$ in $G/N$.$\\[4pt]$
• $T(g)=1\in N$ so $\phi(g+N)=T(g)+N=N$, hence $\phi(g+N)=0$ in $G/N$.

Thus $\phi$ is not injective.

Answer 2

Your mapping is absolutely fine. Let $f: gN\in G/N\mapsto T(g)N\in G/N$.

Then $gN\in\ker(f)$ if and only if $T(g)\in N$, so $\ker(f)=T^{-1}(N)/N$.

Thus $f$ will be injective if and only if $T^{-1}(N)=N$, or equivalently $T(N)=N$.

I suppose that Herstein was a bit sloppy here, and forgot to mention that either $G$ or $N$ is finite (in this case $T(N)\subset N$ is equivalent to $T(N)=N$), or that $T(N)=N$.