# How is $A_3$ normal in $S_3$

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?

• Another way of looking at Normal subgroups is the following: if H is a normal subgroup, then Hx=xH, ie. the left cosets are equal to the right cosets. This however does not imply that H is a Centralizer. Sep 2, 2020 at 22:42
• Normality is akin to commutativity but at the level of subgroups. It is not the same however. Sep 2, 2020 at 22:43
• H being normal does not imply that hc=ch. Sep 2, 2020 at 22:43
• You can also argue that because $[S_3:A_3]=2$, $A_3$ is a subgroup of index $2$, so it is normal in $S_3$. Sep 2, 2020 at 22:48
• @ilovebulbasaur Why is it that a subgroup of index 2 has to be normal? Update: Ah, never mind, it's because it has only two cosets H and Hx, so Hx=xH. Anyway, that's an interesting argument. Sep 2, 2020 at 23:11

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 $$G$$.