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I am ultimately want to prove that $A_{5}$ is simple and the first step in doing so is to:

$(a)$ Write out all partitions of $5.$ Which of these correspond exclusively to even permutations?

I was able to write all partitions of $5.$ But I was not able to say which of these correspond exclusively to even permutations. My idea is to uses this fact first to find which elements of $S_{5}$ corresponds to this partition:

If the partition is $j_1 + \dots + j_k = n$ where $j_1 \geq \dots \geq j_k$, then we can map it uniquely to the permutation $(1 \dots j_1) (j_1 + 1 \dots j_1 + j_2) \dots (n-j_k+1 \dots n)$ (I would appreciate if someone tell me how did we find this function?)

And then try to write each of them as a product of even number of transpositions and for the ones that I will succeed to do so, I will consider them to be the even permutations. Am I correct in this idea or this is a more efficient way of doing so?

Also, I found this question here

Finding the parity of a permutation "exclusively"? and I am wondering what does the word exclusively mean in my question ? does it means unique?

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  • $\begingroup$ Every permutation has a corresponding partition, given by the cycle lengths of the permutation, but many permutations can correspond to the same partition. For example $(123)(45)$ and $(254)(13)$ both correspond to $3+2=5$. The word "exclusively" here means you're looking for partitions such that all of the corresponding permutations are even. (For example, there are 20 permutations corresponding to $3+2=5$, including the two above, and they are all odd.) In practice, the word "exclusively" can be ignored, as permutations with the same cycle structure automatically have the same parity. $\endgroup$
    – Ravi Fernando
    Commented Oct 11, 2020 at 15:23
  • $\begingroup$ @RaviFernando and what about my idea for the solution is it a correct idea? $\endgroup$
    – user777833
    Commented Oct 11, 2020 at 15:25
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    $\begingroup$ Yes, that's one way. As you are likely aware, every permutation can be expressed as the product of either an even number or an odd number of transpositions, but not both. So once you have written a permutation as a product of transpositions, you can immediately read off its parity. Alternatively, the answers to the question you linked may help speed up your calculations. $\endgroup$
    – Ravi Fernando
    Commented Oct 11, 2020 at 15:55
  • $\begingroup$ @RaviFernando thank you! $\endgroup$
    – user777833
    Commented Oct 13, 2020 at 3:29

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