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$\begingroup$

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?

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    Okay thats great thanks! Also thanks alot for the latex tip!2012-03-28

1 Answers 1

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$\{(12)(34),(13)(24)\}$ 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 $N_1$.

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    @aash: The original post uses angle brackets, not curly brackets. Angle brackets are shorthand for "The subgroup generated by..." whatever is in the brackets. That is, $\Bigl\langle (12)(34), (13)(24)\Bigr\rangle \neq \Bigl\{ (12)(34), (13)(24)\Bigr\};$the left hand side is the *subgroup* generated by $(12)(34)$ and $(13)(24)$, which is exactly what you describe as $N_1$.2012-03-29