$D_6$ is not a subset of $D_8$

Question

I came across an example in Chapter-2 of Dummit and Foote(page-47) which says :$D_6$ is not a subgroup of $D_8$ ,the former is not even a subset of latter.I can't understand why is it not the subset of $D_8$? How do we define one group as a subset of another?

Answer 1

The statement "$D_6$ is not a subgroup of $D_8$" means there is no subgroup $H\subset D_8$ such that $D_6$ is isomorphic to $H$. This is easy to verify as $D_6$ has an element of order $3$, but no subgroup of $D_8$ has an element of order $3$.

It is possible, though not particularly useful, to regard $D_6$ as a subset of $D_8$ set-theoretically.

Answer 2

$D_{2n}$ is usually defined as the group of symmetries of the regular $2n$-gon. That is $D_6$ acts on a hexagon, while $D_8$ acts on a octagon. You can write them down explicitly and see why $D_6$ is not a subset of $D_8$.

Answer 3

It appears to be the case that ord $r$ in $D_6$ is 3, while ord $r$ in $D_8$ is 4, that's one reason why $D_6$ is not a subgroup of $D_6$.