For $n>1$ ;

Let $H$ be the set of all permutations in $S_n$ that can be expressed as a product of a multiple of $4$ transpositions .Show that $H=A_n$


We know that $A_n$ is generated by $3-$ cycles .

Now $H\subset A_4$ since every element of $H$ is a product of even number of transpositions.

Now take any $\alpha\in A_n$ then $\alpha$ is a $3$-cycle or a product of $3$ cycles.

If $\alpha$ is a product of $3$ cycles of the form $\alpha=(x_1x_2x_3)\ldots (y_1y_2y_3)$ then $\alpha=(x_1x_3)(x_2x_3)\ldots (y_1y_3)(y_1y_2)\in H$.


If $\alpha$ is a $3-$ cycle then $\alpha=(x_1x_2x_3)$

If $\alpha\in A_3$ and if we suppose $\alpha=(123)$ in $A_3$ then how can we write $\alpha$ as a product of $4$ $2-$ cycles?

Please help.


Since $n=2, 3$ are allowed, I presume there is no requirement that the transpositions should be distinct?

For large enough $n$, if distinctness is required within each four, and for any $n$ if it is not, when you multiply $(a b)(c d)(e f)(g h)$ by $(g h)(e f)(c d)(x y)$ you get $(a b)(x y)$ - an arbitrary product of two transpositions.

Without distinctness you can also write $(a b)(x y)=(a b)(c d)(c d)(x y)$ for an arbitrary $(c d)$

  • $\begingroup$ But you did not say how to write $(123)$ as a product of 4 2-cycles in $A_3$ $\endgroup$ – Learnmore Sep 25 '17 at 4:04
  • $\begingroup$ @Maths_Student Depending on which way you multiply cycles (assuming you start with the right-hand one first) try $(12)(12)(12)(23)$ - but this is not a factorisation in $A_3$ since $A_3$ contains no transpositions. $\endgroup$ – Mark Bennet Sep 25 '17 at 6:19
  • $\begingroup$ So $(123)=(13)(12)=(12)(12)(13)(12)$ just this only $\endgroup$ – Learnmore Sep 25 '17 at 6:26
  • $\begingroup$ You could have also $(12)(13)(12)(13)=(132)^2$ if you want to avoid the obvious cancellation. $\endgroup$ – Mark Bennet Sep 25 '17 at 6:29

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