Checking if this proof of $A_{5}$ is generated by $3$-cycles is a complete proof?

Question

The question I want to answer is:

Show that $A_{5}$ is generated by $3$-cycles.

Here is the solution I found online:

EDIT:

First: Showing that $(ab) = (1b)(1a)(1b).$

$L.H.S = (ab) = \begin{pmatrix} 1 & 2& \dots& a& \dots &b& \dots & n \\ 1 & 2& \dots &b& \dots &a&\dots &n \\ \end{pmatrix}\quad \quad (1)$

Now, $R.H.S$ in terms of permutation notation is:

$R.H.S = (1b)(1a)(1b)$

$= \begin{pmatrix} 1 & 2& \dots& a& \dots &b& \dots & n \\ b & 2& \dots &a& \dots &1&\dots &n \\ \end{pmatrix} \begin{pmatrix} 1 & 2& \dots& a& \dots &b& \dots & n \\ a & 2& \dots &1& \dots &b&\dots &n \\ \end{pmatrix} \begin{pmatrix} 1 & 2& \dots& a& \dots &b& \dots & n \\ b & 2& \dots &a& \dots &1&\dots &n \\ \end{pmatrix}$\

$$ = \begin{pmatrix} 1 & 2& \dots& a& \dots &b& \dots & n \\ 1 & 2& \dots &b& \dots &a&\dots &n \\ \end{pmatrix}\quad \quad (2)$$

Hence from $(1)$ and $2,$ we have that $L.H.S = R.H.S$.

Now starting the main required proof

Let $\sigma \in A_{5}$ be an arbitrary permutation. Since every permutation in $A_{5}$ is an even permutation i.e. consists of an even number of transpositions then $\sigma = \tau_{1} \circ \tau_{2} \circ \dots \circ \tau_{2m}$ for $m \in \mathbb{N}$ and $\tau_{i}$ are all transpositions.

Let $\tau_{1} = (ab), \tau_{2} = (cd),$ then $\tau_{1} = (1b)(1a)(1b) $ and $\tau_{2} = (1d)(1c)(1d).$ And therefore $$\tau_{1} \circ \tau_{2} = (1b)(1a)(1b)(1d)(1c)(1d) = (1 a b)(1 d b)(1 d c) $$ Which is a product of $3$-cycles.

Now, Since $\sigma$ consists of an even number of transpositions, then it is the product of $3$-cycles i.e.it is generated by $3$-cycles and since $\sigma$ was arbitrary permutation of $A_{5},$ then $A_{5}$ is generated by $3$-cycles as required.

And here is a solution I was directed to by some members of this site $A_n$ is generated by 3-cycles given $n\geq 3$. Is this proof correct?

My question is:

Is the first solution is complete (it does not distinguish between disjoint transpositions and not disjoint ones)?

In my opinion, the first solution is simpler and clearer.

Answer 1

You should distinguish the cases where the transpositions are not disjoint.

This is because if in the case that $b=d$, you will get $$\tau_1\tau_2=(1ab)(1bb)(1dc).$$ where in this case $(1bb)=\iota$ which is not a $3$-cycle. So for this case, you have to write $$\tau_1\tau_2=(1ab)(1bc).$$

Answer 2

An even simpler solution. Let $x\in A_5$. Then $x=(i_1,j_1)...(i_{2n},j_{2n})$: product of even number of involutions. So it is enough to prove that every product $(a,b)(c,d)$ is a product of $3$-cyles. If $\{a,b\}$ and $\{c,d\}$ are not disjoint, then the product $(a,b)(c,d)$ is a 3-cycle. So it is enough to consider the case when these sets are disjoint. Notice that $(1,4,2)(3,4,2)=(1,2)(3,4)$ (the first permutation in the product acts first). Conjugating by arbitrary element of $S_5$, we get that any product of two disjoint involutions $(a,b)(c,d)$ is a product of two $3$-cycles (because conjugation preserves the cycle structure), as required.