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if p is prime , show that an element has order p in Sn iff it's cycle decomosition is a product of commuting p-cycles

my solution is very diffrent about the one in the book and I don't know if my strategy is right

my proof

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let T is an element of Sn

and the cycle decomposition of T = t1 t2 t3 .... tm

t1 , t2 , ... , tm are disjoint cycle

so

ti tj = tj ti for all i,j

T^p = ( t1 t2 .... tm )^p = (t1)^p (t2)^p .... (tm)^p

now , ti , tj are 2 disjoint cycle so if ti or tj are not equal to identity permutation then ,

ti tj =\= e

now if T^p = 1

then

(t1)^p (t2)^p .... (tm)^p = 1

but (t1)^p , (t2)^p , ... , (tm)^p are disjoint

so if (ti)^p =\= 1 for all i then

(t1)^p (t2)^p .... (tm)^p =\= 1

but (t1)^p (t2)^p .... (tm)^p = 1

so it's nessary that

(ti)^p = 1 for all i

so

l ti l = p

and ti 's are commuting because they are disjoint

so we proved that order of T = p where p is prime then it's cycle decomposition has the same order p

now

every t1 is a p-cycle

is this right ?

the otherway is easy to prove

plz help

in this question we can't use the rule which says @ order of an element in Sn is the LCD of the lenghts of the cycles in its cycle decomposition because this rule didn't come in the book and I search for a proof with out it