# Prove that every element in a normal subgroup N is also in the conjugate of N

group-theorynormal-subgroups

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 $$N$$.

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