# Every alternating permutation is a product of 3-cycles [duplicate]

Show that every element in $A_n$ (alternating group of degree $n$) for $n \ge 3$ can be expressed as a $3$-cycle or a product of three cycles.

I understand that if $n$ is odd, then any element can be written as (let the numbers represent arbitrary elements) $$(1,2,3)(3,4,5)(5,6,7)...(n-2,n-1,n)$$ as a product of $3-$cycles, but when $n$ is even I can't figure out how to write it as a product of three cycles.

## marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 23 '16 at 19:59

• I don't understand your attempt? How do propose to write for example the element $(12)(34)$ of $A_5$ (so $n$ is odd) as a product of 3-cycles? – Jyrki Lahtonen Jul 23 '16 at 19:53
• Did you even bother to search? Ten seconds with google gave me two exact duplicates on our site, and many other variants! – Jyrki Lahtonen Jul 23 '16 at 20:04

Every element of $A_n$ is a product of an even number of transpositions (=$2$-cycles).

Now note that

$(ab)(cd) = (acb)(acd)$

$(ab)(ad) = (adb)$

Any element of $A_n$ is the product of cycles. Any cycle in $A_n$ may be expressed as a product of an even number of consecutive transpositions. The product of two consecutive transpositions is a three-cycle.

• "The product of two consecutive transpositions is a three-cycle". Only if they are not disjoint. – lhf Jul 23 '16 at 19:52
• Indeed, I must be more precise. By consecutive transpositions, I mean of the form $(ab) (bc)$. – Ashvin Swaminathan Jul 23 '16 at 19:53
• Cf. the comment here. – Benjamin Dickman Jul 23 '16 at 19:53

Suppose you have an even permutation $a_1,a_2\dots a_n$ of the integers $1,2,3\dots n$.

We just have to prove we can sort them back, using only $3$-cycles, we use induction.

When $n=2$ they must be sorted from the start.

So suppose we can do it for $n-1$. After applying cycle $(a_1,1,x)$ (where $x$ is any other element) we can observe the last $n-1$ elements form an even permutation on $n-1$ objects, so we can sort them by the inductive hypothesis.

• Did you even bother to search? Ten seconds with google gave me two exact duplicates on our site, and many other variants! – Jyrki Lahtonen Jul 23 '16 at 20:04
• Was this intended for OP? – Jorge Fernández Hidalgo Jul 23 '16 at 20:06
• No. To you. After all, experienced answerers are better placed to suspect that a question is such a standard result/exercise that it has surely been done on the site earlier. – Jyrki Lahtonen Jul 23 '16 at 20:16
• Oh, I see, my bad, – Jorge Fernández Hidalgo Jul 23 '16 at 20:22
• Don't worry about it too much. Just something to consider. I belong to the more militant wing of the anti-duplicate party. I don't know exactly where the majority of our users lie. Judging from the discussions on meta there are users like me, others who don't care at all, and a lot of folks in between (as you might expect). – Jyrki Lahtonen Jul 23 '16 at 20:40