Given the following subgroup of $S_6$: $G=\langle(1,2,3),(1,2,4),(5,6)\rangle$ I'm asked to show that $H= \langle (1,2,3),(1,2,4)\rangle$ is normal on G and isomorphic to $A_4$ and also to find the order, center and derived/commutator subgroup of G.
I think I got the first part right: since (5,6) is disjoint with the other two it commutes with them, so I can easily see that conjugating H with any element of G stays in H. Also, H generates all the 3-cycles that move $(1,2,3,4)$, so $H=A_4$ (http://people.math.gatech.edu/~ecroot/3cycle.pdf)
I'm having trouble with the rest. By Lagrange's theorem I know $12 \mid |G| \mid 720=|S_6| $, but I don't know the exact value.
I have also found that $(5,6)\in Z(G)$ since it commutes with the 3 generators, but I don't know whether there are more elements in Z(G) and I don't know how to find the derived.
Best Answer
You're off to a good start:
Great, so indeed $H$ is normal in $G$.
You might want to substantiate this claim, depending on what you have seen or shown so far, of course. Then it does follow that $H\cong A_4$. From here it is not hard to prove that $$G=H\times\langle(5\ 6)\rangle\cong A_4\times S_2,$$ see also this question. With this characterization of $G$, finding the order, center and commutator subgroup of $G$ is a matter of finding the orders, centers and commutator subgroups of $A_4$ and $S_2$.