Non-normal Subgroup of Quaternion Product

Question

I needed to show that the $Q_8\times Q_8$ contains a non-normal subgroup where $Q_8$ is the quaternion group.

My example was the subgroup $\{(1,1), (i,k), (-i,-k)\}$.

I got the question wrong because, according to Lagrange's theorem, my subgroup cannot actually be a group because its order does not divide the order of $Q_8 \times Q_8$.

This makes sense; however, I do not understand why this set fails to be a group explicitly. It seems to me that it is closed and associative. It obviously has the identity and the other 2 elements are inverses of each other.

If this set is not a group it should fail one of these criteria, unless there is something I am not understanding.

Would anyone be able to shed some light on why my set fails to be a group explicitly other than Lagrange's Theorem?

Answer 1

Each subgroup $H$ of a group $G$ must be closed under every product in $H$, including powers of elements. Since

$$(i,k)^2=(i^2,k^2)=(-1,-1)$$

is not in your example, it cannot be a subgroup.