How to show that the set of orthogonal n x n matrices forms a group under multiplication I am studying orthogonal matrices and I am not sure if to show if a set of orthogonal $n \times n$ matrices forms a group under multiplication. We must check each of the group axioms.
I found that the axioms are:

Closure
Associativity
Existence of identity matrix
Existence of the inverse matrix.

I edited my question, since I was able to find more information about this topic.
This group is called $O(n)$.
To check the four axioms I did:
Let $A \text{ and } B \in O(n)$, denoted as orthogonal matrices and assume that $C=AB$, then,
Closure :

To prove that $C \in  O(n)$ we must prove that $C$ is a real $n \times n$ orthogonal matrix with uni-modular determinant. Since A and B are real $n \times n$ matrices, $C$ is also a real $n \times n$ matrix so,
$C^TC=(AB)^T AB=B^T A^T AB = B^TB=I$

Associativity :

Matrix multiplication is associative, so the law holds for $O(n)$ group elements.

I am not sure if this is enough to prove associativity.
Identity element :

The $n \times n$ identity matrix $I_{n \times n}$ represents the identity element.

In this case I am not sure if this is enough to prove the identity element.
Inverse element :

Let $A^{-1}$ be the inverse of $A$, then we need to prove that  $A^{-1} \in  O(n)$ since $(A^{-1})^T=(A^T)^{-1}$. We have that:
$(A^{-1})^T A^{-1}=(A^T)^{-1} A ^{-1}=(AA^T)^{-1}=I^{-1}=I$

Can anyone check if what I did is correct? I also would like to know if I can prove the associativity and the identity element in a better way.
thanks
$(U_1 U_2)^T (U_1 U_2) = I$, hence $U_1 \circ U_2$ is orthogonal.

Associativity follows from associativity of matrix multiplication.

The matrix $I$ is an identity for matrix multiplication.

$U^T U = U U^T = I$, hence $U^{-1} = U^T$ is the required inverse.
Your doubt about the identity element is correct.

First you need to show that $I \in O(n)$ by showing that
$$I^TI=II^T=I,~~\text{since}~I^T=I.$$ 
Then
$$AI=IA=A, \forall A\in O(n).$$