# Why must a finite symmetry group be discrete?

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

• What do you think? What would one that isn't discrete look like? – vonbrand Feb 25 '13 at 4:37
• Where is the symmetry group getting its topology from exactly? In a somewhat recent question, it was seen that any finite topological group is discrete modulo a normal subgroup N (which is in particular the connected component of the identity), meaning the cosets in G/N form a topological basis. Thus there can be topologies on finite groups that are not quite discrete. – anon Feb 25 '13 at 4:39
• I know an infinite symmetry group would describe a figure like a circle since there are infinitely many rotations but I don't quite understand how a finite symmetry group has to be discrete. – Peter12 Feb 25 '13 at 4:43
• It doesn't have to be discrete (only discrete modulo the identity component), unless perhaps the finite symmetry group is getting its topology in some particular manner from how it acts or there are extra restrictions on the topology. – anon Feb 25 '13 at 4:45
• Peter, what does it mean to you to say a group is discrete? What does it mean to say a group is not discrete? – Gerry Myerson Feb 25 '13 at 4:55

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 $|U \cap U'| = |U| - 1$.