Why must a finite symmetry group be discrete?

Question

I'm having trouble justifying why a finite symmetry group is discrete. Can someone help?

Answer 1

Consider a finite symmetry group, $G=S_n$ (I'm assuming that you meant the full symmetry group, otherwise, as was pointed out in the comments, the question seems to be not well-defined).

A topology $\mathcal T$ on $G$ would be a subset of $2^G$, satisfying the axioms of a topological space. A topology is discrete if any point is open, or equivalently $\mathcal T = 2^G$. A topology is trivial if $\mathcal T = \{\emptyset, G\}$. Since any set can have the trivial topology, we will need to assume that $\mathcal T$ is not trivial.

In a topological group a translation by a group element is a homeomorphism, so if U is open then so is $gU$ for all $g \in G$.

Lets verify that $\mathcal T$ is indeed discrete. Because it is not trivial there is some open set $U \in \mathcal T$, such that $U\neq G$ and $U \neq \emptyset$. Lets say that $|U|=k$ with $0<k<n$. Observe that as $g$ runs over $G$, $\{gU\}$ consists of all possible subsets of $\{1...n\}$ of size $k$, and since the union of open sets is open, we get that any $V \subset G$ with $k\le |V| \le n$ is open.

To see that every set with less than $k$ elements is open as well, we use the fact that the intersection of two open sets is open, and prove by downward induction on the size of smallest non-trivial open set. Namely, given $x\in U, y \notin U$ define $U' = U \setminus \{x\} \cup \{y\}$ which is also open, and note that $|U \cap U'| = |U| - 1$.

I hope this helps.