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Let P be a partition of a group G with the property that for any pair of elements A, B of the partition, the product set AB is contained entirely within another element C of the partition. Let N be the element of P that contains 1. Prove that N is a normal subgroup of G and that P is the set of its cosets.

This exercise was asked about in the following questions: partition of a group to have normal subgroup partition of a group and cosets Proving a partition is set of cosets Artin 2.10.3 help understanding why $AN=NA=A$ implies $N$ is normal

My question is about Brian Bi's proof linked here, where it is claimed that $1 \in P_n$.

The following is a screenshot of the proof (Kiefer Sutherland's voice):

enter image description here

Please explain the $1 \in P_n$. This is the only part I don't understand.

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    $\begingroup$ $P_n$ is the partition to which $n$ belongs, i.e., $N$. But we already know that $1\in N$ $\endgroup$ – user418131 Oct 15 '18 at 11:05
  • $\begingroup$ @AnotherJohnDoe $N=P_n=P_1$? Thank you! $\endgroup$ – user198044 Oct 15 '18 at 11:09
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    $\begingroup$ Yes, that's right $\endgroup$ – user418131 Oct 15 '18 at 11:10
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$n$ belongs to some element $P_n$ of the partition $P$.

$n$ belongs to $N$.

We are given that $N$ is not just any subset of $G$: $N$ is also an element of $P$.

$\therefore, N \cap P_n \ne \emptyset \implies N = P_n$

$1 \in P_1 = N \implies \therefore, 1 \in P_n$.

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