Argue that real matrices which are orthogonal form a group Argue that real matrices which are orthogonal form a group with matrix multiplication as the composition.
I know that a group have to satisfy $3$ axioms.
$1.$ $$(a \circ b) \circ c = a \circ (b \circ c)$$
$2.$ $$e \circ a = a \circ e =a$$ where $e$ is identity element
$3.$ $$a \circ a^{-1}=a^{-1} \circ a= e$$
But how should I begin? And is it a finite group?
 A: *

*It is clear from the definition of multiplication between matrices;

*Take $e = I_n$ the identity matrix (which is orthogonal);

*If $A$ is orthogonal, then $A \cdot A^T = I$, therefore $A^{-1} = A^T$. Meaning to say that every element in the set (usually denoted by $\mathcal O(n)$) has an inverse. 
To conclude you need to show that the set of all orthogonal matrices is closed under multiplication. Well, if $A,B$ are orthogonal then $$(A\cdot B)\cdot(A\cdot B)^T = (A \cdot B) (B^T\cdot A^T) = A \cdot (B\cdot B^T) \cdot A^T = \ldots$$ 
I believe you can take it from here
A: As noted in the comments the key fact here is to prove that the set of orthogonal matrices is closed under multiplication. So, let $A$, $B$ orthogonal matrices, i.e. (by definition) such that $AA^T=I$ and $BB^T=I$. we have:
$$
AB(AB)^T=ABB^TA^T=AA^T=I
$$
so also $AB$ is orthogonal.
Now the other properties are obvious:
1) come from distributivity of matrix multiplication.
2) is true because the identity matrix is orthogonal
3) is proved since  $AA^T=I \Rightarrow A^TA=I \Rightarrow A^-1=A^T$ .
