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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!

Soof_fie
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3 Answers3

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We know every 3-cycle in $S_n$ is in $A_n$, because $(a\,b\,c)=(a\,c)(a\,b)$, and by definition $A_n$ consists of everything that can be written as a product of an even number of transpositions.

Thus, in particular, every product of $k$ 3-cycles is a product of $2k$ transpositions and therefore in $A_n$. So whatever the 3-cycles generate must be a subgroup of $A_n$ too.

hmakholm left over Monica
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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, $A_n\subseteq \langle \sigma_1,\cdots, \sigma_s\rangle$

user2345678
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Every 3-cycle is an even permutation, and any product of even permutations is an even purmtation. So, any product of 3-cycles is in $A_n$, which is the subgroup of all even permutations.

Lee Mosher
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