Any hint on : Every $A_{n}$ elemnt is $n$-cycles product.

Asked Jul 28 '15 at 16:46

Active Feb 25 '20 at 00:40

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[Added explanation] I found this exercise as follows in Hungerford : Abstract algebra (3rd edition) page 236, exercise number 40. Stated as follows :

C.40. Prove that every element of $A_{n}$ is a product of $n$-cycles.

$n$-cycle is explained in the book like the permutation : $(1 2 3 4 ....n)$

[Original text]

Every $A_{n}$ element is a product of $n$-cycles . Tried everything i could imagine, though got further, could not find a solution How could i prove it? A hint is what i am looking for. $A_{n}$ : The Alternating group , the subgroup of the group of permutations $S_{n}$ in a set of $n$ elements.

edited Jul 28 '15 at 17:37

asked Jul 28 '15 at 16:46

pigeon

1

What you wrote is not clear. Do you mean, "Every element of $A_n$ is a product of $n$-cycles"? – Zev Chonoles Jul 28 '15 at 16:48
see this – Jack's wasted life Jul 28 '15 at 17:06
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@ZevChonoles That wouldn't make sense when $n$ is even because $n$-cycles are not in $A_n$. I haven't a clue what is being asked! – Derek Holt Jul 28 '15 at 17:07
But a product of $n$-cycles might just be in $A_{n}$. Like $(1234)(1234)=(13)(24)$ is in $A_{n}$, but $(1234)$ is not. – pigeon Jul 28 '15 at 17:10
Yes that's true! Well the $n$-cycles generate a normal subgroup of $S_n$, so you could use the fact that for $n \ge 5$ the only such subgroups are $1$, $A_n$, and $S_n$. – Derek Holt Jul 28 '15 at 17:12
Very good start, but "normal group" , and your fact are not known to me , yet. – pigeon Jul 28 '15 at 17:18
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If you multiply $(1,2,3,\ldots,n)$ by the inverse of $(2,1,3,\ldots,n)$ then you get a $3$-cycle. Does that help? – Derek Holt Jul 28 '15 at 18:06
..And $3$-cycles generates $A_{n}$. Perfect. Case closed ! – pigeon Jul 28 '15 at 18:17

Any hint on : Every $A_{n}$ elemnt is $n$-cycles product.

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