Can anyone help with the following?
Prove that G cannot have a subgroup H with the order of H=n-1, where the order of G=n, and n is greater than 2.
Thanks!
We have that $\displaystyle G-H=\{m\}$. Now, clearly $\displaystyle m\ne e$. Now, since $\displaystyle n>2$ we see that $\displaystyle |H|>1$ and so there exists some $\displaystyle h\in H-\{e\}$. Clearly $\displaystyle mh\in G$ but $\displaystyle mh\notin H$ since if it were we'd have that $\displaystyle mh=h',\text{ }h'\in H$ and so $\displaystyle m=h'h^{-1}\implies m\in H$. Thus, $\displaystyle mh\notin H\implies mh=m\implies h=e$. This contradicts our choice of $\displaystyle h$. The conclusion follows.