Isn't this $f$ always a group isomorphism

Question

Consider the following exercise from a book I'm reading:

If $n$ is odd show that $$ f: O(n) \to SO(n) \times \{1,-1\}, A \mapsto (A \operatorname{det}{A}, \operatorname{det}{A})$$

is an isomomorphism.

But why does $n$ have to be odd? I can easily show that it's a group homo., injective and surjective without using any information about $n$. What am I missing here?

Answer 1

If $n$ is even, $n = 2k$, then

$$\operatorname{sgn}(\det(A \det A)) = \operatorname{sgn}(\det A (\det A)^{2k}) = \operatorname{sgn}(\det A) \cdot \operatorname{sgn}(\det A)^{2k} = \operatorname{sgn}(\det A)$$

so $A \det A$ isn't a member of $SO(n)$ unless $A$ was originally.

Answer 2

If $n$ is even, multiplying $A\in O(n)\backslash SO(n)$ by $\det(A)=-1$ should not give a matrix in $SO(n)$.