How is $A_3$ normal in $S_3$

Question

I notice that conjugation works, but:

$(12)(123) = (123)(12) \rightarrow (1)(23)=(13)(2)$

Is clearly wrong. Doesn't normality imply that $hg=gh$ for $H$ being normal in G? What am i doing wrong here?

Answer 1

A normal subgroup H of G, is a subgroup such that for any $h$ from $H$, $xhx^{-1} \in H$ i.e. all the conjugates ($xhx^{-1}$) of $H$ are in $H$. Another way to look at it is the following: the left and right cosets of $H$ are the same, meaning $Hx=xH$. Basically saying that multiplying $H$ by an element is commutative. (Remember $Hx$ is the set all of elements of $H$ multiplied by $x$).

However this does not imply that the elements of $H$ commute with themselves or with other elements of $G$. There is another subgroup which is called the centralizer, by definition it's elements are those elements of $G$ which commute with all of the elements of $G$.