Must subgroups sharing a common element be nested in each other? 
Let $H$ and $K$ be subgroups of a group $G$ which have a common element besides the identity. Does this mean that either $H$ or $K$ is a subgroup of other? 

Let $a$ be the common element, then both subgroups contain the subgroup generated by $a$. I know it is possible for a subgroup to have an element which is not in other subgroup. Any counterexamples?
 A: In the quaternion group, $\langle i\rangle$ and $\langle j\rangle$ intersect in $\{1,-1\}$. 
The other groups with eight elements also have counter-examples, except the cyclic group.
A: Vector spaces are also groups, with addition as the group operation.
Subgroups correspond to subspaces.
In the context of vector spaces, your question is:

Let $H$ and $K$ be subspaces of a vector space $G$ which have a common element besides the origin. Does this mean that either $H$ or $K$ is a subspace of the other?

The problem is now more geometrical than algebraic, and perhaps your intuition will work better.
The answer is no: Let $G=\mathbb R^3$ and $H$ and $K$ two different two dimensional subspaces.
Since your original claim fails for the special case of vector spaces, it is false.
Linear algebra is sometimes a good way to check abstract algebraic ideas.
Notice, however, that vector spaces are always abelian as groups, and not even all abelian groups are vector spaces.
A: Alternatively (because free groups can be a bit hard to picture), let $G$ be $(\mathbb Z,+)$, and let $H$ and $K$ be the subgroups $2\mathbb Z$ and $3\mathbb Z$, respectively. Then $H$ and $K$ have all multiples of $6$ in common, but neither of them is a subgroup of the other.
A: Let $G$ be the free group on three generators. So we can have
$$
G = \langle a,b,c\rangle
\qquad
H = \langle a,b\rangle
\qquad
K = \langle a,c\rangle
$$
A: Consider a Rubik's cube with all its possible configurations with top and right side rotations or top and left side rotations.
Or, simpler, any at least $4$-element sequence with permutations $(2,1,3,4)$ and $(3,2,1,4)$ and with permutations $(2,1,3,4)$ and $(4,2,3,1)$.
