I found the following proposition in "Algebra – Michael Artin (P.59)":
If a group $G$ has just one subgroup $H$ of order $r$, then that subgroup is normal.
I'm having a difficulty understanding how could a group have only $1$ subgroup of order $r$. Because we know from Lagrange's theorem:
Let $H$ be a subgroup of a finite group $G.$ The order of $H$ divides the order of $G$,
. . . that $r$ divides the order of $G$. Since the cosets of this subgroup partition $G$, and all have the same order $r$, therefore $G$ has many subgroups of order $r$. (Their number is equal to how many cosets we have.)
The proposition is easy to see if $r=|G|$ or $r=1$, but otherwise I don't see how. Any help is appreciated.
Best Answer
The mistake made here is confusing cosets with subgroups. It's true we get many cosets of order $r$, but they are not necessarily subgroups. These cosets can be obtained by a subgroup $H$ as follows:
$$aH=\{ah \mid h\in H\}$$
but these cosets are not subgroups. (Because the identity is only in one of these cosets, namely $H$.)