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Let $ G=S_{5} $ . Consider the set L of subgroups of G that are isomorphic to the non-cyclic group of order 4. Let us call two subgroups H and K belonging to L as equivalent if they are conjugate (that is gH$g^{-}$=K for some $g\in G$) .How many equivalence classes are there in L?

My attemp: The equivalence relation on L is induced by the conjugation action of G on L . Let $ H\in L $ then $ H=\{ e, (ab), (cd), (ab)(cd)\} $ or $ H=\{e , (ab)(cd), (ac)(bd), (ad)(bc) \} $ where a,b,c,d are distint and belong to $ \{1,2,3,4,5 \} $.

These are not in same class since they have different cycle types . So, no. of equivalence classes is at least 2 . I am totally confused after this. How to show it is exactly two? ( ans is 2)

Also under the group action $ stab(H)=N_{G}(H) $ for H$\in L$. Now, H is normal in sylow 2 subgroup of order 8 but the normalizer of H could be larger. How to determine it's size for a subgroup H ? (I am trying to use orbit stabiliser thrm)

Are my thinkings correct ? Please help me.

user-492177
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1 Answers1

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Type (i).

If $a,b,c,d$ are distinct and $a',b',c',d'$ are distinct then there is an element of $S_5$ sending $a\mapsto a'$ etc, and this will conjugate $H=\langle (ab),(cd)\rangle $ to $H'=\langle (a'b'),(c'd')\rangle $.

Type (ii).

You say you want to use the Orbit-Stabilizer Theorem. Well $H=\{e, (12)(34), (13)(24), (14)(23)\}$ is clearly normal in $S_4$. As $S_4$ has prime index 5 in $S_5$, the normalizer is either $S_4$ or $S_5$. As $H$ is not normal in all of $S_5$ (it is moved by $(15)$ for example), the normaliser is $S_4$. So there are $5$ conjugates of this $H$. Clearly there are exactly $5$ of type (ii), one for each of the five letters we can omit when constructing the subgroup. So they are all conjugate.

ancient mathematician
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