For a discrete topological space $X$, is Perm$(X)$ a topological group as a subspace of product topological space $X^X$? Let $X$ be a discrete topological space and consider the product space $X^X$ (which is the space of all functions from $X$ to $X$) with product topology. Let $S(X):= \{ f: X \to X : f $ is bijection $\}$ . Then $S(X)$ is a subset of $X^X$ and $S(X)$ forms a group under the operation of function composition. My question is : Under the subspace topology inherited from $X^X$, does $S(X)$ form a topological group ?
 A: Yes.
If $X$ is discrete, then the product topology on $X^X$ agrees with the compact-open topology, see Compact open topology on space of continuous maps from a discrete space is a product.
If function spaces are endowed with the compact-open topology, then it is well-known that the composition function $T : B^A \times C^B \to C^A, T(f,g) = g \circ f$ is continuous provided $B$ is locally compact. Discrete spaces are locally compact, so that $T : X^X \times X^X \to X^X$ is continuous, and therefore also the restriction to $S(X)$. Inversion on $S(X)$ can be treated by directly considering the topology of pointwise convergence: If you take a net $(f_\alpha)_{\alpha \in A}$ of bijections converging pointwise to some $f \in S(X)$, then $(f^{-1}_\alpha)$ converges pointwise to $f^{-1}$. This is true because for all $x \in X$ there exists $\alpha_0$ such that $f_\alpha(x) = f(x)$ for $\alpha \ge \alpha_0$ due to discreteness of $X$. Note that $A$ is any directed partially ordered set.
A: You can also prove this directly.
First, you should write down what does the continuity of the composition operation mean in your case:
For every neighborhood $W$ of $g ∘ f$ there is a neighborhood $U$ of $f$ and a neighborhood $V$ of $g$ such that $g' ∘ f' ∈ W$ for every $f' ∈ U$ and $g' ∈ V$.
It is enough to consider subbbasic neighborhoods $W$ – those when you prescribe a nontrivial open set at one coordinate $x ∈ X$. Since $X$ is discrete, $\{g(f(x)\}$ is open, so we want to find $U$, $V$ such that $g'(f'(x)) = g(f(x))$ for every $f' ∈ U$, $g' ∈ V$. Do you see how to choose $U$ and $V$?
The situation with the continuity of inverse is similar.
