Isn't it possible that $D_4$ has also a subgroup with $1$ element??

Question

A consequence of the Lagrange theorem:

Let $G$ a finite group and $H$ a subgroup of G. Then $|H| \mid |G|$.

is that each subgroup $\neq <i_d>$ of $D_4$, which has $8$ elements , has either $2$ or $4$ elements..

But.... $1 \text{ divides also }8$..Isn't it possible that $D_4$ has also a subgroup with $1$ element??

Answer 1

Of course $D_4$ has one and only one element of order $1$, which forms the trivial subgroup containing only the identity element of $D_4$, and so the subgroup $|\langle id\rangle$ is its unique subgroup of order $1$.

Every group, if it is a group, has a unique identity element, and hence, has one and only one element (and subgroup) of order $1$.

Answer 2

$D_4$ has a subgroup with one element as does every group: the trivial group $\langle id\rangle$.