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Let $A : |A| = k$, and let elements of the power set $P(A)$ be (partially) ordered according to inclusion (that is, $\forall X, Y \in P(A) : X \leq Y \iff X \subseteq Y $). The question is then how many automorphisms exist on $P(A)$.

I think the answer is $k!$, and the sketch of my reasoning is roughly as follows:

  1. Any automorphism on $P(A)$ must map a set of size $l$ to a set of size $l$.
  2. For any automorphism $f$, its part restricted to sets of size $k-1$ defines the rest of the mapping: indeed, let $Z : Z \in P(A), |Z| = k - 2$. Then it's easy to see that there exist just two sets $X_1, X_2$ of size $k-1$ such that $Z \leq X_1, Z \leq X_2$. Since $f$ is an automorphism, $f(Z) \leq f(X_1), f(Z) \leq f(X_2)$, and it's easy to see that the set included in both $f(X_1)$ and $f(X_2)$ is unique, so $f(Z)$ is defined merely by $f(X_1)$ and $f(X_2)$. Applying this recursively proves the claim.
  3. Any permutation of subsets of size $k-1$ is a proper automorphism, once the rest of it is defined in accordance with the previous point.
  4. There are $k$ subsets of size $k-1$.
  5. The $k!$ answer immediately follows.

I have two questions:

  1. Is this reasoning correct?
  2. The factorial suggests there might be a beautiful recursive proof. Is there one?
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    $\begingroup$ I think it's even easier than this. Every permutation of $A$ gives you a permutation of $P(A)$ that preserves all the boolean set operations. You know there are $k!$ permutations of $A$ - perhaps you proved that by recursion. $\endgroup$ – Ethan Bolker Feb 25 '18 at 0:12
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    $\begingroup$ It might be clearer to do this by proving an automorphism is determined by its action on the subsets of cardinality 1. $\endgroup$ – saulspatz Feb 25 '18 at 0:15
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What seems simpler to me is to use that any automorphism on $P(A)$ must map a set of size 1 to a set of size 1, so it permutes the size 1 subsets. This induces a permutation on the elements of $A$ itself. Conversely, any permutation of the elements of $A$ induces an automorphism. We are therefore getting a bijection between permutations of $A$ (of which there are $n!$) and automorphisms of $P(A)$.

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  • $\begingroup$ Right, that (along with the answers given in comments) is definitely more straightforward! Thanks! $\endgroup$ – 0xd34df00d Feb 25 '18 at 0:18

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