Show that a group of order 7 has no subgroups of order 4. I can do this with the use of Lagrange's theorem but my professor says its possible without it. I can't find how to go about solving it. Any hints would be appreciated.

Okay, suppose your subgroup is $G=\{1,a,b,c\}$, and your big group is $H=G\cup \{d,e,f\}$.
Now, what can $ad$ be?  Not $1$, since $a^{-1}\in G$ and $d\notin G$.  Not $a$, since $d\neq 1$.  Not $b$ (resp. $c$) since $a^{-1}b\in G$ (resp. $a^{-1}c\in G$).  Also not $d$, since $a\neq 1$.   Hence $ad\in \{e,f\}$.  By similar logic, $bd, cd\in \{e,f\}$.  By the pigeonhole principle, two of these must agree.  Without loss, suppose $ad=e=bd$.  But now $add^{-1}=bdd^{-1}$ so $a=b$, a contradiction.
