Checking the normality for finite versus infinite subgroups of a group My question has been touched upon already here and here.
Dummit and Foote give a flurry of exercises (3.1.27 to 3.1.29) starting with 'Let $N$ be a finite subgroup of a group $G$.'
The first of these asks to show that $gNg^{-1} \subseteq N$ iff $gNg^{-1}=N.$ As I understand the previous posts, if N is finite, a counting argument can be used. Do I understand well that this is something along the following lines? 
Assume $gNg^{-1} \subseteq N$. One can devise a map $f$ from $N$ to $gNg^{-1}$ as $f(n)=gng^{-1}$. Since this map is injective, it also holds that $N \subseteq gNg^{-1}$, proving the equality. Would this reasoning not hold in the infinite case?
Sincerely, I do not see why this argument would be simpler than the following, which also hold in the infinite case: since $gNg^{-1} \subseteq N$ for all $g \in G$, it holds as well that $g^{-1} Ng\subseteq N$, and hence $N \subseteq  gNg^{-1}$.
However, D&F continue the exercises for finite $N$. In brief, they ask to prove that normality of $N$ in $G$ can be checked from the generators of $N$ and of $G$.
So, my question is whether this would also hold for infinite $N$; I do not see why it would not. But, then, why do D&F insist of finiteness of $N$? This puzzles me all the more, because a preceding exercise (3.1.26.c) did not require finiteness of $N$.
 A: You are incorrect in concluding that $N\subseteq gNg^{-1}$ follows from the fact that the map $n\mapsto gng^{-1}$ is injective. I mean, the map $[0,1]\to[2,3]$ given by $x\mapsto x+2$ is injective, but that doesn't tell me that $[0,1]\subseteq [2,3]$. 
Rather, the fact that the map is injective tells you that $|N|\leq |gNg^{-1}|$ (in fact, you always get equality, because the map is also onto). Since $gNg^{-1}\subseteq N$, you also get $|gNg^{-1}|\leq|N|$, so now you have $|gNg^{-1}|=|N|$. In the finite case, since $gNg^{-1}\subseteq N$, the fact that they have the same cardinality allows you to conclude that $gNg^{-1}=N$.
In the infinite case, this is no longer true. And in fact, it is possible for $gNg^{-1}\subseteq N$ to hold for every $g$ in a generating set, and yet for $N$ to not be normal because the inclusion is proper.
For example, if $G$ is the free group in two letters $x$ and $y$, and $H$ is the group generated by all elements of the form $uxu^{-1}$, where $u$ is a positive word (a group elements written as a product of $x$s and $y$s where all the exponents are nonnegative), then $xHx^{-1}\subseteq H$ and $yHy^{-1}\subseteq H$, $\{x,y\}$ is a generating set for $G$, but $y^{-1}xy\notin H$, so $H$ is not normal. The problem here is that even though $|yHy^{-1}|=|H|$ and $yHy^{-1}\subseteq H$, you still have $yHy^{-1}\neq H$. 

Consider the following four statements:

  
*
  
*For all $g\in G$, $gNg^{-1}=N$.
  
*For all $g\in G$, $gNg^{-1}\subseteq N$.
  
*$\langle S\rangle = G$, and for all $s\in S$, $sNs^{-1}=N$.
  
*$\langle S\rangle = G$ and for all $s\in S$, $sNs^{-1}\subseteq N$.
  

Turns out that statements 1, 2, and 3 are equivalent, so they all imply normality (statement 1). However, statement 4 is weaker in the general case (though equivalent when $N$ is finite).
Clearly 1 implies the remaining three, 2 implies 4, and 3 implies 4. 
You've noticed already that 2 implies 1: using the given statement with $g^{-1}$ instead of $g$, we get $g^{-1}N(g^{-1})^{-1} \subseteq N$, which gives $g^{-1}Ng\subseteq N$ for any $g\in G$; then multiplying on the left by $g$ and on the right by $g^{-1}$, we conclude that for all $g\in G$, $N\subseteq gNg^{-1}$; together with 2, this yields condition 1.
To see that 3 implies 1, first note that for every $s\in S$ we get $N=s^{-1}Ns$ by taking $sNs^{-1}=N$ and multiplying on the left by $s^{-1}$ and on the right by $s$; fromt his, given an arbitrary $g\in G$, write it as a product of elements of $S$ and their inverses and use 3 to conclude that $gNg^{-1}=N$.
The problem for 4, however, is that you can't get that $s^{-1}Ns\subseteq N$ for $s\in S$ from the property given. So you can never make the jump to either 2 or 3 in the general case. It is only when $N$ is finite (or $S$ is closed under inverses) that you can get the equality you need to push the argument through.
