I've been been working through Socratica's video on normal subgroups and factor groups and they show how $y^{-1}Ny \subseteq N$ for some normal subgroup $N \trianglelefteq G$ and $y \in G$. They leave it as an exercise to show $y^{-1}Ny$ and $N$ are equal by demonstrating that $N \subseteq y^{-1}Ny$. I tried starting this proof by picking some $n_1 \cdot y^{-1} \in Ny^{-1}$ and $n_2 \cdot y \in Ny$, then stating that their product is in $Ny^{-1}y$. Then

$(n_1 \cdot y^{-1}) \cdot (n_2 \cdot y) = n_3 \cdot y^{-1} \cdot y$

$n_1 \cdot (y^{-1} \cdot n_2 \cdot y) = n_3 $

$n_1^{-1} \cdot n_1 \cdot (y^{-1} \cdot n_2 \cdot y) = n_1^{-1} \cdot n_3 $

$(y^{-1} \cdot n_2 \cdot y) = n_4$

To me, the last line reads that for any element in $N$ there is a corresponding element in $y^{-1}Ny$. But I'm not sure if I'm any closer to proving it since I'm pretty sure that last line is something we could state just by defining $n_2$ and $n_4$ as elements in $N$.

## Best Answer

You need to start with an element $n_1 \in N$

$$ n_1 = y^{-1} y n_1 y^{-1} y\\ $$

Let $x=y^{-1}$ and use that $x^{-1} N x \subseteq N$ because the statement you had was for all $y \in G$ including $x$.

$$ y n_1 y^{-1} = x^{-1} n_1 x\\ = n_2\\ $$

$$ n_1 = y^{-1} n_2 y \\ n_1 \in y^{-1} N y $$