# Section 8.2 Part 2: Analysis

While this comment could have been absorbed into the previous comment on this section, it is important to note that Hungerford specifies that $Na=aN$ does not imply that $na=an$ for every $n\in N$ when dealing with groups and normal groups.

I find it interesting that the statement $a^-1 Na\subseteq N$ for every $a\in G$, where $aNa^-1 ={ana^-1\mid n\in N$ and the statement $a^-1 Na=N$ for every $a\in G$ are equivalent.