The following question is from 'Matrix Groups for Undergraduates' by Kristopher Tapp:
Let $G \subset GL_n(\mathbb R)$ be a compact subgroup.
(1) Prove that every element of $G$ has determinant $1$ or $-1$.
(2) Must it be true that $G \subset O(n)?$
Hint: Consider conjugates of $O(n).$
Attempt: I know the relevant definition of closed (i.e. if a sequence of matrices in $G$ has a limit in $GL_n(\mathbb R)$, then that limit must lie in $G$). I also know a matrix element is bounded if (when it is regarded as an element of $\mathbb R^{n^2}$) it has finite norm. I just cannot really see how these give us any information about the determinant.
As for $(2)$, following the hint and considering a conjugate of $O(n)$, when we consider $A \in O(n)$ and $B \in GL_n(\mathbb R)$ we see that \begin{equation} \begin{split} \det BAB^{-1} &= (\det B)(\det A)(\det B^{-1}) \\ &= (\pm 1)(\det B)(\det B^{-1}) \\ &= \pm 1 \end{split} \end{equation}
so we see that conjugates are in the group, implying it is a normal subgroup, but what does this tell us about G?
I'm definitely missing a lot here, any help would be appreciated, thanks all.
If $G$ is compact, then $\det(G)$ is a compact subgroup of $(\mathbb{R},\times)$. There are only two such subgroups: $\{1\}$ and $\{\pm1\}$.
For the other question, consider the group$$\left(\left\{\begin{pmatrix}1&0\\0&1\end{pmatrix},\begin{pmatrix}1&-2\\0&-1\end{pmatrix}\right\},.\right).$$It is compact, but not a subgroup of $O(2,\mathbb{R})$.