Just while reading, I saw that,

" For each $X≠∅$ subset of $N$ , $ (P(X), Δ)$ is subgroup of $(P(N), Δ)$"

I know all these notations, I mean, I know, the power set of natural numbers forms group with respect to the operation "symmetric difference Δ.

"My question is", why they had taken $X≠ ∅$ ? What if $X= ∅$ ? Is $(P(∅), Δ)$ is subgroup of $(P(N), Δ)$"? Or not?

My attempt: clearly $P(∅)= \{∅\}$ is nonempty subset of $P(N)$.

Further as,

$∅Δ∅ = (∅-∅) ∪ (∅-∅)$ $ = ∅∈ P(∅)$ Hence, $P(∅)$ is non empty, finite subset of $P(N)$, which is closed under the same binary operation as in $(P(N),Δ)$. Hence $(P(∅), Δ)$ is non-trivial subgroup of $(P(N), Δ)$.

Is am I right?


1 Answer 1


Yes, you are.

If you want to be technical about it, you need to check that the set $P(\emptyset)$ is closed not only under the multiplication but also the inversion. For example, $(\mathbb N, +) \subseteq (\mathbb Z, +)$ is closed under addition but not under negation, and thus not a subgroup.

However, each element in the group you are considering has finite order, so that doesn't matter.


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