# Let $P$ be a partition of a group with $AB \subseteq C$. Why is $1 \in P_n$? $P_n$ is the equivalence class of $n \in N$ and $1 \in N=P_1$.

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.

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):

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

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

$$n$$ 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 $$P$$.
$$\therefore, N \cap P_n \ne \emptyset \implies N = P_n$$
$$1 \in P_1 = N \implies \therefore, 1 \in P_n$$.