What is the difference between left- and right-conjugation in a group?

Question

A subgroup $N$ of a group $G$ is called normal if $gNg^{-1} = N$ for all $g\in G$. As I understand, for a fixed $n\in N$, the map $g\mapsto gng^{-1}$ is called left conjugation of $n$ by $g$. Now a priori, I have no reason to believe that this is always equal to the map $g\mapsto g^{-1}ng$, which we would call right conjugation of $n$ by $g$. For $G$ need not be abelian, so it could be that $g^{-1}ng\ne gng^{-1}$. Could we replace the condition $gNg^{-1} = N$ with $g^{-1}Ng=N$ in the definition of a normal subgroup? If so, is it just convention to use "right" conjugation?

Answer 1

As mentioned, the definition of normal subgroup specifies that

$g Ng^{-1} = N \tag 1$

for all $g \in G$. We note this entails

$\forall g \in G, \; g^{-1}N g = g^{-1}N(g^{-1})^{-1} = N; \tag 2$

therefore, though the mappings

$g \mapsto gng^{-1} \tag 3$

and

$g \mapsto g^{-1}ng \tag 4$

are not in general the same, they give rise to the same definition of normal subgroup, since conjugation by every $g \in G$ is required; thus we always pick up both left and right conjugation by any paricular $g$. So in fact the choice of right or left conjugation is arbitrary when defining normal subgroup.

In closing, note that left and right conjugation are the same, that is,

$gng^{-1} = g^{-1}ng \tag 5$

if and only if

$g^2n = ng^2, \; \forall n \in N, \; g \in G; \tag 6$

i.e., when ever every element of $N$ commutes with every square in $G$, when the set of squares is contained in $C_N(G)$, the centralizer of $N$ in $G$.

Answer 2

If we set $h=g^{-1}$ it transforms $gNg^{-1}$ into $h^{-1}Nh$ so it's really an arbitrary choice and we just pick one.