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So it is not hard to see that the conjugacy class of $(123)(456)$ in $S_7$ has cardinality $\frac{7!}{3\cdot 3\cdot2}=280$.

But for instance to go from $(123)(456)$ to $(132)(456)$ you have to conjugate by $(23)$ which is not in $A_7$ so $(132)(456)$ is not in our class (call it $C$). Same for $(123)(465)$. But $(132)(465)\in C$ because $(23)(56)\in A_7$.

I could probably find all the elements of $C$ by trying out different permutations but I feel like there is a better way...

We know that $|A_7|={|S_7|\over2}$ and conjugating $(123)(456)$ by any permutation in $S_7\supset A_7$ gives a permutation of the type $(abc)(def)~$ with $a,...,f\in\{1,...,7\}$ and $a,b,...,f$ are disjoint because otherwise the "permutation" by which we conjugate wouldn't be bijective.

So I'm tempted to say that $|C|=|\bigcup\limits_{~a\in A_7}\{(~a(1)a(2)a(3)~)~(~a(4)a(5)a(6)~)~\}|=\frac{280}{2}$ but that's probably false.

A hint would be appreciated

John Cataldo
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  • Hint: Can you show that if $C$ is a conjugacy class in $S_{n}$ that is a subset of $A_{n}$, then it is either a conjugacy class in $A_{n}$ or it splits into a union of two $A_{n}$ conjugacy classes. – Adam Higgins Jan 06 '19 at 11:27

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Hint: If you conjugate $(123)(456)$ by $(14)(25)(36)$, an odd permutation, you get $(123)(456)$ back. Therefore you also get $(132)(456)$ from $(123)(456)$ by conjugating it by an even permutation.

Jyrki Lahtonen
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  • This theme comes up frequently on our site, hence only a hint in a community wiki post. This if probably one of the highest voted threads getting to, if not the bottom of the matter, then at least relatively close. There are many other good duplicate targets. – Jyrki Lahtonen Jan 06 '19 at 11:27