I am revising for my group theory exam and am trying to work the composition series for the Alternating Group $A_4$. I think if we let;
$$N_1 = \, \langle(1 2)(3 4)\,,\, (1 3)(2 4)\rangle \quad\text{and}\quad \hspace{7pt} N_2 = \, \langle (1 3)(2 4)\rangle$$ then we have a composition series
$ 1 \, \unlhd \, N_2 \unlhd \, N_1 \,\unlhd \,A_4$.
But I am not sure what the other ones are? How many are there in total?
Best Answer
$\{(12)(34),(13)(24)\}$ is not a subgroup of $A_4$ because it does not contain identity element.
I think $N_1$ should be $$N_1=\{1,(12)(34),(13)(24),(14)(23)\}$$ Note that $N_1$ is a normal subgroup of $A_4$.
Now take $N_2=\{1,(12)(34)\}$ or $\{1,(13)(24)\}$ or $\{1,(14)(23)\}$
These are normal in $N_1$ because they are of index $2$ in $N_1$.