I haven't studied group theory in earnest beyond first courses, so my notation may be nonstandard and my question may be a 'standard fact', so bear with me:
Consider a group $G$, and for each natural number $n \in \mathbb{N}$ define $G_n := \{g \in G : g^n = 1\}$ to be the set of $n^\text{th}$ roots of unity in $G$. If $G$ is Abelian, then each $G_n$ is a subgroup of $G$, but in general this may not be the case: for example, taking $G = D_3 = \langle \sigma, \tau : \sigma^2 = \tau^3 = 1, \tau \sigma = \sigma \tau^2 \rangle,$ we can compute that $ G_2 = \{1,\sigma,\sigma \tau, \sigma \tau^2\}$ which is not a subgroup of $D_3$. This leads to the first question, which I've already answered:
Are there any non-Abelian groups $G$ such that every $G_n$ is a subgroup of $G$?
The smallest such $G$ is the quaternion group, assuming my calculations are correct. So we refine this question:
Which non-Abelian groups $G$ are such that every $G_n$ is a subgroup of $G$?
I haven't made any calculations for groups larger than the quaternion group, but I thought I'd throw this question out into the open. However, the calculation above for $D_3$ shows that no dihedral groups $D_n$ with $n \geq 3$ satisfy this property, since $(D_n)_2 = \{1,\sigma,\sigma \tau, \ldots, \sigma \tau^{n-1}\}$ is not a subgroup of $D_n$.
Have at it!