find the centre of S_n

**abhishekkgp** · March 5th 2011, 08:17 PM

question 1) Prove that $Z(S_n)=1 \text{ for all } n \geq 3$ .

question 2) Prove that if $H \trianglelefteq G$ with $|G:H|=p,\text{ p is a prime}$ , then for all subgroups K of G either $K \leq H$ or,
$G=HK \text{ and } |K:K \cap H|=p$ .

I dont know how to solve the first one but i could solve the second one.
here is my solution of the second one:

since $H \trianglelefteq G$ we find that $HK = KH$ so $HK \leq G$ .
also $H \trianglelefteq HK$ and $HK/H \leq G/H$ so it makes sense to consider $|G/H:HK/H|$ .

Using $|HK|= (|H||K|)/|H \cap K|$ we get,

$|G/H:HK/H| = |G:H||H|/|HK| = p|H \cap K|/|K| = p/|K: H \cap K|$ .

This means either $|K: H \cap K|=1 \text{ or } p$
this easily lads to the desired result.

If you have a different solution to the second one then please post it.

**Drexel28** · March 6th 2011, 10:32 AM

Originally Posted by abhishekkgp

question 1) Prove that $Z(S_n)=1 \text{ for all } n \geq 3$ .

question 2) Prove that if $H \trianglelefteq G$ with $|G:H|=p,\text{ p is a prime}$ , then for all subgroups K of G either $K \leq H$ or,
$G=HK \text{ and } |K:K \cap H|=p$ .

I dont know how to solve the first one but i could solve the second one.
here is my solution of the second one:

since $H \trianglelefteq G$ we find that $HK = KH$ so $HK \leq G$ .
also $H \trianglelefteq HK$ and $HK/H \leq G/H$ so it makes sense to consider $|G/H:HK/H|$ .

Using $|HK|= (|H||K|)/|H \cap K|$ we get,

$|G/H:HK/H| = |G:H||H|/|HK| = p|H \cap K|/|K| = p/|K: H \cap K|$ .

This means either $|K: H \cap K|=1 \text{ or } p$
this easily lads to the desired result.

If you have a different solution to the second one then please post it.

For the first one....don't think too hard. Given any element of $S_n\;\; n\geqslant 3$ just construct something which doesn't commute with it. There are two more ways I can see to prove the second one, but the one you used is easiest.

**abhishekkgp** · March 6th 2011, 10:43 PM

Originally Posted by Drexel28

For the first one....don't think too hard. Given any element of $S_n\;\; n\geqslant 3$ just construct something which doesn't commute with it. There are two more ways I can see to prove the second one, but the one you used is easiest.

i got the solution to the first one after i posted it.
can you post your solutions of the second one???

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