Is SU(2) a subgroup of the exceptional lie group $G_2$? I am not an expert in Lie groups and I have spent ages looking at textbooks; I assume that because I haven't found this statement explicitly it must either be untrue or obvious ;)
The only thing I can find is that $G_2$ definitely has a subgroup $SU(3)$*, and I know (?) that $SU(3)$ has a subgroup $SU(2)$ - surely that means the answer to my question is yes? Any help gladly appreciated!
Added: And if the answer is yes, is there an 'obvious' way to see it, possibly in terms of the octonions themselves? (This is in reference to the algebra, not the gorup - but I have noticed that any three elements of the (imaginary) octonions themselves form a Lie algebra isomorphic to $\mathfrak{su}(2)$ (since they are alternating), is this related?)
*Source http://arxiv.org/pdf/0902.0431.pdf
 A: It is a basic fact of group theory that if $H$ is a subgroup of $K$ and $K$ is a subgroup of $G$ then $H$ is a subgroup of $G$. This should be a straightforward exercise encountered right after the definition of subgroups, which is way before learning about Lie groups or octonions. So if $SU(2)$ is a subgroup of $SU(3)$ and $SU(3)$ is a subgroup of $G_2$, then of course $SU(2)$ is a subgroup of $G_2$.
Here, we can identify $G_2$ with the automorphism group of the octonions $\mathbb{O}$. Then $SU(3)$ is stabilizer of $i$, or equivalently the pointwise stabilizer of $\mathbb{C}$, making the orthogonal complement within $\mathbb{O}$ a copy of $\mathbb{C}^3$. And $SU(2)\cong Sp(1)$ is the pointwise stabilizer of the quaternions $\mathbb{H}$, making the orthogonal complement of $\mathbb{H}$ within $\mathbb{O}$ another copy of $\mathbb{H}$ which $Sp(1)=S^3$ acts on effectively by multiplication.
Any imaginary octonion generates a copy of $\mathbb{C}$, any two linearly independent ones generate a copy of $\mathbb{H}$, and any three "algebraically" independent imaginaries (i.e. none of the three are contained in the subalgebra contained by the other two) generate all of $\mathbb{O}$. That implies the "subalgebra" generated by commutators from two independent imaginaries is a copy of $\mathfrak{so}(2)$, and from three independent imaginaries is $(\mathbb{R}^7,\times)$ where $\times$ is the 7D cross product induces from octonion multiplication. This is not a Lie algebra; the failure of octonions to be associative implies the Jacobi identity is not satisfied.
