Prove that $gNg^{-1} \subseteq N$ iff $gNg^{-1} = N$ I'm reading the solution to one of my homework problems and am stuck on something. Here is the problem:


  Let $N \leq G$ be a finite subgroup. Show that $gNg^{-1} \subseteq N$
  if and only if $gNg^{-1} = N$.


The solution is to suppose that $gNg^{-1} \subseteq N$ and consider the map $\varphi: N \to gNg^{-1}$ defined by $\varphi(n) = gng^{-1}$. Observe that such a map is a bijection and thus $|N| = |gNg^{-1}|$. The equality then follows.

I'm just curious as to why we can assume that $\varphi$ is a bijection (i.e. why can't there exist $n_1, \, n_2 \in N$ such that $gn_1g^{-1} = gn_2g^{-1}$ where $n_1 \neq n_2$)?
The fact that $\varphi$ is a bijection follows from the invertibility of group multiplication.

Start with $gn_1g^{-1} = gn_2g^{-1}$. First multiply both sides on the left by $g^{-1}$.  Then multiply on the right by $g$.
One implication : $gNg^{-1} = N $ implies $gNg^{-1} \subset N $ is trivial. Lets try reverse part:
Suppose $gNg^{-1} \subset N $  and we will show that $gNg^{-1} = N $.
On contrary suppose that $gNg^{-1} \neq N $
so there is $ n_1 \in N $ such that
$gng^{-1} \neq n_1$  for all $n \in N$ then on pre and post multiplication by $g^{-1}$ and g we obtain
$ n \neq g^{-1}n_1g $ for all $ n \in N $
which in turn says that $ n \notin  g^{-1}Ng$  for all $ n \in N$ contradicting our hypothesis. This contradiction proves the result.