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?

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 $a\in G$.