How many equivalence classes are there in $J$.?

Question

Let $G$ be the symmetric group $S_5$ of permutations of five symbols. Consider the set $J$ 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 $J$ as equivalent if they are conjugate (that is, there exists $g $ such that $gHg^{-1}=K$). How many equivalence classes are there in $J$.?

I have tried a lot but i found only one. But correct answer is two. But how ?

Answer 1

Here's a hint: in one class, all the elements of order $2$ are double transpositions, in the other class, there are single transpositions.

Answer 2

EDIT: Originally, this answer made the mistaken assumption that the involutions that generate $V$ were transpositions in $S_5$.

The Klein four-group is generated by a pair of commuting transpositions involutions $s$ and $t$: $$ V = \langle s, t \rangle = \{ 1, s, t, st \}. $$

There are two cases to consider.

Class 1: pair of commuting transpositions and a double transposition. An embedding of $V \hookrightarrow S_5$ is determined by selecting two non-overlapping pairs $\{a,b\}$, $\{c,d\}$. The subgroup consists of $$ 1, \quad (a\,b), \quad (c\,d), \quad (a\,b)(c\,d). $$ But this double-counts the possible embeddings since the order you choose generating transpositions doesn't matter. Thus, the total number is $$ \frac{1}{2}\binom{5}{2} \binom{3}{2} = \frac{1}{2} \cdot 10 \cdot 3 = 15. $$

Class 2: three double transpositions. An embedding of $V \hookrightarrow S_5$ is determined by selecting a set of four elements $\{a,b,c,d\}$. The subgroup consists of $$ 1, \quad (a\,b)(c\,d), \quad (a\,c)(b\,d), \quad (a\,d)(b\,c). $$ Since all possible pairs of pairs are in the subgroup and any two can be generators, the count is exact and the total number is $$ \binom{5}{4} = 5. $$

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All $15$ embedded copies of $V$ in Case 1 are conjugate, and all $5$ embedded copies of $V$ in Case 2 are conjugate, but embedded copies of $V$ from opposite classes are not conjugate. Why? Any two permutations are conjugate by some $g \in S_5$ if and only if they are of the same cycle shape. Explicitly, using cycle notation, $$ g \, (a\,b) \, g^{-1} = (a'\,b') \qquad \text{where } g(a) = a' \text{ and } g(b) = b' $$ for Class 1, and $$ g \, (a\,b)(c\,d) \, g^{-1} = (a'\,b')(c'\,d') \qquad \text{where } g(a) = a', g(b) = b', g(c) = c', \text{ and } g(d) = d' $$ for Class 2.