Define a normal subgroup of G $N$ is a normal subgroup of $G$ if $aNa^{-1}$ is a subset of $N$ for all elements $a $ contained in $G$. Assume, $aNa^{-1} = \{ana^{-1}|n \in N\}$.
Prove that in that case $aNa^{-1}= N.$
If $x$ is in $N$ and $N$ is a normal subgroup of $G$, for any element $g$ in $G$,  $gxg^{-1}$ is in $G$. Suppose $x$ is in $N$, and $y=axa^{-1}$ as is defined. Since $N$ is normal, $aNa^{-1}$ is a subset of N. 
$x= a^{-1}ya$. Given that $x$ is in $N$, and $x=a^{-1}ya$, $y$ is also in $N$. If $y$ is in N, then $axa^{-1}$ is also in $N$. $X$ is in $aNa^{-1}$.
Does the proof make sense?
 A: I'm a little confused as to how you got to your last step. Here's my version.

Suppose that $aNa^{-1} \subseteq N$ for all $a \in G$. Then to show that equality holds, it suffices to show that $bNb^{-1} \supseteq N$ for all $b \in G$. To this end, choose any $b \in G$ and choose any $x \in N$. We want to show that $x \in bNb^{-1}$.
Now recall that $x \in N$ and that $aNa^{-1} \subseteq N$ for all $a \in G$. Thus, by taking $a = b^{-1}$, it follows that:
$$
(b^{-1})x(b^{-1})^{-1} = b^{-1}xb \in b^{-1}Nb \subseteq N
$$
so that $b^{-1}xb \in N$. But then we're done, since:
$$
x = exe = (bb^{-1})x(bb^{-1}) = b\underbrace{(b^{-1}xb)}_{\in ~ N}b^{-1} \in bNb^{-1}
$$
as desired. $~~\blacksquare$
A: Here's another way to see where the result comes from. If $N$ is normal,
$$\begin{align*}
aNa^{-1}&\subset N\\
aN&\subset Na.
\end{align*}$$
But the definition is symmetric in $a$! Swapping the roles of $a$ and $a^{-1}$, we also get $Na\subset aN$. Thus $aN=Na$, which gives you $aNa^{-1}=N$.
Philosophical aside: Groups can be annoying to work with when elements don't commute. However, oftentimes the next best thing is knowing that certain subgroups commute. That's what it means to be normal: you commute with all elements. All the important properties of normal subgroups (especially the formation of quotients) follow from this observation.
A: By definition, we have $aNa^{-1}\subseteq N$ for all $a\in G$. This is equivalent to saying $aN\subseteq Na$ for all $a\in G$. (Why?) 
Thus, we simply need to show $Na\subseteq aN$ for all $a\in G$.
To this end, take $x\in Na$. Then $x=n_1a$ for some $n_1\in N$, and so $a^{-1}x=a^{-1}n_1a=(an_2a^{-1})^{-1}$. (What should $n_2$ be for this to make sense? Why can I do this?)
Now, because $N$ is normal, $an_2a^{-1}\in N$. But this also implies $(an_2a^{-1})^{-1}\in N$ (Why?) and therefore $a^{-1}x\in N$. Hence, we can write $a^{-1}x=n_3$ for some $n_3\in N$. So $x=an_3$ meaning $x\in aN$. 
Hence we have $Na\subseteq aN$, and thus $Na=aN$; or, $aNa^{-1}=N$. However, our choice of $a$ was arbitrary so we have proven our claim for all $a\in G$.
