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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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    $\LaTeX$ tip: use `\langle` and `\rangle`, not `<` and `>`.2012-03-28
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    Sure: you could take $N_2 = \langle (12)(34)\rangle$, or $N_2 = \langle (14)(23)\rangle$. $A_4$ has no subgroups of order $6$, and no subgroup of order $3$ is normal, so those are all of them.2012-03-28
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    Okay thats great thanks! Also thanks alot for the latex tip!2012-03-28

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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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    Hi. I have beautified this post and fixed it free of spelling errors and unjustifiable use of capital letters. [This](http://math.stackexchange.com/editing-help) might be useful to tell you how you could emphasize relevant parts of your answer and edit them in general. And, please look at the way site is run. In particular, we do not encourage users to post in SMS language. Regards,2012-03-29
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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