# Need help for the example on conjugation.

This is an example from Dummit & Foote text, i've some queries in this-

If $|G|>1$, then unlike action by left multiplication,$G$ does not act tranistively on itself by conjugation because {1} is always a conjugacy class (i.e., an orbit for the action).More generally,the one element subset {a} is a conjugacy class iff $gag^{-1}=a$ and for all $g \in G$ iff $a$ is in the centre of $G$.

My questions are-

• Does here $|G|>1$ refers to order of non-abelian group?
• What is meant by "unlike action by left multiplication"?What if here 'unlike' is replaced by 'like' ?

$|G|$ refers to the order of $G$ (abelian or not). Action by left multiplication is always transitive (consequence of the so-called 'sudoku property' of groups). The conjugation action however, is not transitive for any non-trivial group $G$.
• It refers to the left group action defined by the group's multiplication operation. A group action is a map $G\times X \rightarrow X$ satisfying certain properties. In this case the set $X$ is none other than the set of the group $G$, and the action of a $g\in G$ on an $x \in X$ is none other than $g.x=g\star x$. (the action of $g$ on $x$ is the same as taking the product in $G$ of $g$ and $x$ with $g$ on the left). – Justin Benfield Jun 26 '16 at 10:18
• There is, but whenever you have a right group action, there is an equivalent left group action, hence you can considerably simplify notation, conventions, and language, by sticking exclusively with left actions. The map that expresses the equivalence is any anti-automorphism from $G$ to $G^{op}$ from the group $G$ to it's opposite group $G^{op}$ (e.g. the group theory example here: en.wikipedia.org/wiki/Antihomomorphism). Edit: also mentioned here: en.wikipedia.org/wiki/Group_action – Justin Benfield Jun 26 '16 at 10:31