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Let H={e,(12)(34)} and K ={e, (12)(34), (13)(24), (14)(23)} be subgroups of $ S_4$ where e is identity element. Then which of following is true

  1. H and K are normal subgroups of $S_4$

  2. H is normal in K and K is normal in$ A_4$

  3. H is normal in $A_4$, but not normal in $ S_4$

  4. K is normal in $S_4$ but H is not

How should this question be approached, checking left and right cosets of these subgroup is very time consuming. Checking gh$ g^{-1}$ $\in $ H for every g in $ S_4$ is also time consuming.

Arnaud D.
  • 20,884
Foggy
  • 1,259

1 Answers1

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Two elements in $S_n$ are conjugated if and only if they have the same factoriziation to disjoint cycels.

Hence $K$ is normal in $S_4$ and therefore in $A_4$.

From the same reason $H$ is not normal in $S_4$.

$H$ is normal in $K$ since it is a subgroup of index $2$.

I left for you to cheack if $H$ is normal in $A_4$.

Ofir Schnabel
  • 3,891
  • If elements are conjugated, why would k be normal? – Foggy Jan 29 '15 at 09:00
  • @Foggy , let $k=(ab)(cd)$ where ${a,b,c,d}={1,2,3,4}$ be an element in $S_4$ and let $g$ be another element. Then $g^{-1}kg=(xy)(zw)$ where ${x,y,z,w}={1,2,3,4}$. Since all the elements of the form $(xy)(zw)$ are in $K$ it is stable under conjugation$=$ normal. – Ofir Schnabel Jan 29 '15 at 09:04
  • This rule holds for S_n, and A_n also? – Foggy Jan 29 '15 at 09:06
  • Yes and no.. the difference is in $S_n$ any two elements are conjugated if and only if they have the same factorization to cycels. In $A_n$ you have just the "only if" part. That is, if two elements in $A_n$ are conjugated then they have that same factorization to cycels. The converse is not true. – Ofir Schnabel Jan 29 '15 at 09:10