For my course in Group Theory, I have seen various proofs that show why the alternating group $A_n$, which consists of the elements of $S_n$ that can be expressed as an even number of transpositions (i.e. 2-cycles), is generated by the 3-cycles.
All of these proofs, and sometimes also the question, seem to guide you to showing that any element in $A_n$ can be expressed as a product of 3-cycles. Now I get the proofs up to this point.
What I do not understand, and I hope you can help me with, is why the fact that any element in $A_n$ can be expressed as a product of 3-cycles means that $A_n$ is generated by the 3-cycles. Could it not be that, even though any element of $A_n$ can be expressed as a product of 3-cycles, that if we let the 3-cycles generate a group there will be elements in that group that are not in $A_n$? I do not see why our proof (for instance given here) would exclude that possibility.
If any of you could shed some light on this, your help is very much appreciated!
Best Answer
I don't think that both resolutions given are complete.
Let $\sigma_1,\cdots, \sigma_s$ be the 3-cycles from $S_n$. From the given answers, it was shown that $\langle \sigma_1,\cdots, \sigma_s\rangle \subseteq A_n.$ It remains, then, to show that $A_n \subseteq \langle \sigma_1,\cdots, \sigma_s\rangle. $
Let $\alpha \in A_n$. We know that $\alpha$ can be written as a product of transpositions, and by the parity of $\alpha$ it must be the product of an even number of transpositions. Now note that the product of two transpositions is always a product of 3-cycles: indeed, if $\tau_1 = (a_1,a_2), \tau_2 = (b_1,b_2)$ are disjoint, then $\tau_1\tau_2 = (a_1b_1a_2)(b_1b_2a_1);$ and if they have an element in common, say $a_2=b_1$, then $\tau_1\tau_2 = (b_1b_2a_1).$ Now, we've shown that $\alpha$ has an even number of transpositions and since the product of two transposition is a 3-cycle, $\alpha$ is then a product of 3-cycles. Hence, $A_n\subseteq \langle \sigma_1,\cdots, \sigma_s\rangle$