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I know what conjugacy classes are. However, you deal with two groups ##A_5## and ##S_5## and neither said where ##g## is from nor what ##x## is. This is needed to unambiguously speak of conjugacy classes. Sorry, but my crystal ball is out of order. My guess is that you talk about ##S_5##, for otherwise you won't get ##24## elements in a class, but why?The conjugacy classes are defined the usual way in group theory: Conjugacy class - Wikipedia. For a group G and element x in it, a conjugacy class is the set of all values of g.x.g^{-1}for all g in G.

A cycle is ##(n_1 \,\ldots \,n_k)## with one length ##k##. Now what are cycles in a cycle? And how can there be more than one length?The cycle content is the set of lengths of cycles in it.

You wrote e.g. "5:even 24" and in the legend "(cycle content): (parity) (number of elements)"The number of elements is for the conjugacy class.

Now does this mean we have ##24## ##5-##cycles, or that each conjugacy class of one ##5-##cycle contains ##24## elements? I assume the latter, as you just said the number of elements in a class.

So you mean if ##\pi = \sigma_1 \ldots \sigma_k## is a presentation of ##\pi## with disjoint cycles ##\sigma_i##, then you inspect the set ##\{\,\operatorname{ord}(\sigma_i)\,|\,1 \leq i \leq k\,\}##? And the set, not the ordered vector ##(\operatorname{ord}(\sigma_i))_i##, is that correct?Sharing a set of cycle lengths means all the elements having the same set of cycle lengths.

Now ##1⋅2^2:even 15## means: All conjugacy classes of a product of two transpositions have ##15## elements? And you could have omitted all the ##1^k## as they are the identity anyway? Why don't you say this instead of a cryptic "##1⋅2^2:even 15##" code.

Does this mean you considered ##|gA_5g^{-1}|## for ##g\in S_5##? What for? Both classes of ##A_5## have sixty elements.There is a theorem about permutations that conjugation of a permutation does not change its cycle lengths.]Symmetric group - Wikipedia and Alternating group - Wikipedia both discuss the contents of the groups' conjugacy classes. For the symmetric group, "two elements of Snare conjugate in Snif and only if they consist of the same number of disjoint cycles of the same lengths."

I brought in S5 because I wanted to derive the sizes of the classes of A5.

Thanks, I have a proof. I only try to understand yours. You do not need to explain standard vocables, you need to use them correctly (conjugacy classes without the source of conjugation and element), and explain your inventions: set of cycle lengths, cycles contain cycles.An is Simple contains my argument about class sizes and normal subgroups near the bottom of the second page.