Find the number of elements needed to generate a dense subset of $SO_4(\mathbb{R})$ Consider the topological group $G=SO_4(\mathbb{R})$
Find the smallest $n \in \mathbb{N}$ such that $\exists s=\{g_1,g_2...g_n\} \subset G$ for which $<s>$ is dense in G.
$<s>$ denotes the subgroup of G generated by S.
 A: To save the labor, I will write $SO_4$ for $\operatorname{SO}_4$ and write $so_4$ for $\mathfrak{so}_4$.

The answer is $2$.
We first note that a general element of $SO_4(\Bbb R)$ is a double rotation, i.e. if $g \in SO_4(\Bbb R)$, then there is an orthogonal decomposition $\Bbb R^4 = V \oplus W$, such that $g$ is the composition of a rotation on $V$ and a rotation on $W$.
This means that a single element $g$ cannot generate a dense subgroup of $SO_4(\Bbb R)$, because $g^n(V) \subseteq V$ for any $n\in \Bbb Z$, and (since $V$ is closed) hence $h(V)\subseteq V$ for any $h$ in the closure of the subgroup generated by $g$.

Now it suffices to show that two elements can generate a dense subgroup of $SO_4(\Bbb R)$.
It is easier to work with the Lie algebra. Recall that the Lie algebra $so_4(\Bbb R)$ is $6$-dimensional, with basis $(A_i, B_i)_{i = 1, 2, 3}$, as defined here. Below is a screen shot of the matrices:

The Lie bracket is given, for $\{i, j, k\} = \{1, 2, 3\}$, by:


*

*$[A_i, A_i] = [A_i, B_i] = [B_i, B_i] = 0$;

*$[A_i, A_j] = \pm A_k$;

*$[A_i, B_j] = \pm B_k$;

*$[B_i, B_j] = \pm A_k$.


The signs $\pm$ are not quite important for us.
Now pick any two numbers $\lambda, \mu\in \Bbb R$ such that $2\pi, \lambda, \mu$ are linearly independent over $\Bbb Q$. Let $X$ be the matrix
$$\begin{pmatrix}
\cos\mu & 0 & 0 & -\sin\mu\\
0 & \cos\lambda & -\sin\lambda & 0\\
0 & \sin\lambda & \cos\lambda & 0\\
\sin\mu & 0 & 0 & \cos\mu
\end{pmatrix}.$$
In other words, $X$ is a double rotation with respect to the decomposition $\Bbb R^4 = V\oplus W$, where $V = \Bbb Re_2 \oplus \Bbb Re_3$ and $W = \Bbb Re_1 \oplus \Bbb Re_4$.
If we look at the group $H$ of all double rotations with respect to the decomposition $V\oplus W$, then it is clearly isomorphic to $(\Bbb R/2\pi \Bbb Z)^2$ with the obvious isomorphism, and our element $X$ corresponds to the element $(\lambda, \mu)$ under this isomorphism. By our choice of $\lambda, \mu$, we see that $X$ generates a dense subgroup of $H$.
Now pick another matrix $Y$ as follows
$$\begin{pmatrix}
\cos\lambda & 0 & -\sin\lambda & 0\\
0 & 0 & 0 & 0\\
\sin\lambda & 0 & \cos\lambda & 0\\
0 & 0 & 0 & 0
\end{pmatrix}.$$
The number $\lambda$ could be the same as before, or could be any irrational real number. For similar reason, $Y$ generates a dense subgroup of $K$, which is all rotations on the plane $\Bbb R_1 \oplus \Bbb R_3$.
We now prove that $X$ and $Y$ generates a dense subgroup of $SO_4(\Bbb R)$.
Let $G$ be the closure of the subgroup generated by $X$ and $Y$. Then $G$ is a closed subgroup of $SO_4(\Bbb R)$, and hence its Lie algebra $g$ is a Lie subalgebra of $so_4(\Bbb R)$.
But we already see that $G$ contains $H$ and $K$, and hence $g$ contains the corresponding Lie subalgebras $h$ and $k$.
It is clear that $h$ has dimension $2$, with generators $A_1$ and $B_1$; and $k$ has dimension $1$, with generator $A_2$. From the Lie bracket formulas, we see that these three generate the whole $so_4(\Bbb R)$.
Hence $g$ is the whole $so_4(\Bbb R)$ and we are done.
