Is the external direct product of $O(2)$ with itself isomorphic to $O(2)$ i.e.
$O(2) \oplus O(2) = O(2)$
I'm guessing they're not, as to prove that is was, I'd have to define an explicit isomorphism between the groups and we are rarely asked to do that.
But I'm having trouble with this problem showing that they're not isomorphic. I know that all the matrices in $O(2)$ represent either reflections or rotations in $\mathbb{R^2}$.
I suppose if I were to take the direct product (composition in this case) of two reflections, two rotations or a reflection and a rotation, I would just end up with some rotation again.
Is this a reasonable argument to conclude that the two groups are not isomorphic? $O(2) \oplus O(2)$ contains only rotations, where as $O(2)$ has rotations but reflections as well. So they cannot be isomorphic.
Any help would be appreciated. Thank you kindly.
Here's a quick way to argue that these groups are not isomorphic: $O(2) \oplus O(2)$ contains a subgroup isomorphic to $C_2^4$ ($C_2$ denotes the cyclic group of order $2$), but $O(2)$ contains no such subgroup.
I will leave it to you to find an example of such a subgroup in $O(2) \oplus O(2)$.
On the other hand, here's one way to prove that $O(2)$ contains no such subgroup. Suppose that $A_1,A_2,A_3,A_4 \in O(2)$ generate a subgroup isomorphic to $C_2^4$. That is, these matrices commute, each satisfy $A_i^2 = I$, and satisfy no other relations; in particular, $A_i \neq I$ for any $i$.
Because these $A_i$ satisfy $A_i^2 = I$, they must be (orthogonal) reflections, and so they must be symmetric. A (finite) collection of commuting symmetric matrices must be simultaneously orthogonally diagonalizable, so up to an inner automorphism, these four matrices must be of the form $$ \pmatrix{\pm 1 & 0\\0 & \pm 1}. $$ There are only four matrices of this form, one of which is the identity matrix. But this contradicts our earlier statement that $A_i \neq I$ for each $i$.