# 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?

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.

• And also the very same example in the finite cyclic group $G = \mathbb{Z}/12\mathbb{Z}$ (like a clock), where $H$ and $K$ are (additively) generated by $\hat{2}$ and $\hat{3}$ respectively. (Hats are just the members (residue classes) of $\mathbb{Z}/12\mathbb{Z}$.) Explicitly, $H=\{ \hat{0},\hat{2},\hat{4},\hat{6},\hat{8},\hat{10} \}$ and $K=\{ \hat{0},\hat{3},\hat{6},\hat{9} \}$ with $H \cap K =\{ \hat{0},\hat{6} \}$. Then neither $H \subset K$, nor $K \subset H$. – Jeppe Stig Nielsen Oct 18 '16 at 9:47
• Thanks. But should the finite group example be necessarily cyclic? – jnyan Oct 18 '16 at 12:50
• @jnyan: Not necessarily, but cyclic groups are simple and nice-behaved, so they can be argued to make especially striking counterexamples. (You can take the $C_{12}$ example and make a direct product with any group, however wild or complex, you like -- and then this counterexample will lift to your product group, as preimages under the projection homomorphism). – hmakholm left over Monica Oct 18 '16 at 13:02

Let $G$ be the free group on three generators. So we can have $$G = \langle a,b,c\rangle \qquad H = \langle a,b\rangle \qquad K = \langle a,c\rangle$$

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.

• The smallest cyclic group to hold a counterexample would be the cyclic group of order $12$ (my comment to Henning Makholm's answer). – Jeppe Stig Nielsen Oct 18 '16 at 9:58
• @JeppeStigNielsen On the other hand, finding examples is not too hard in general, as a finite group which is not a counter example is almost cyclic of prime power order. – Tobias Kildetoft Oct 18 '16 at 10:59
• @TobiasKildetoft The cyclic group of order $15$ does not seem to contain a counterexample, yet $15$ is no prime power. The requirement is that $H \cap K$ is neither trivial nor equal to $H$ or $K$. – Jeppe Stig Nielsen Oct 18 '16 at 11:13
• @JeppeStigNielsen Yeah, my "almost" was rather imprecise (and I have not actually worked through how to make it precise, though I think these groups should be classifiable). – Tobias Kildetoft Oct 18 '16 at 11:15
• Don't the Klein four-group and more generally $(\mathbb{Z}/p\mathbb{Z}) \times (\mathbb{Z}/p\mathbb{Z})$ constitute abelian non-cyclic groups which are not counterexamples? I am considering creating a new thread with this problem. – Jeppe Stig Nielsen Oct 18 '16 at 12:33

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.

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 $(4,2,3,1)$.