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.
 A: Let $H$ be a hypothetical subgroup of order 4 and suppose $x\notin H$. Then $xH$ cannot be disjoint from $H$ because the group has only 7 elements. Thus there are $a,b\in H$ such that $xa=b$. But then $x=ba^{-1}\notin H$, contradicting the assumption that $H$ is a subgroup.
A: Let $G$ be a group of order $7$ and $H$ a subgroup of order $4$, say $H=\{e,x,y,z\}$. Take $a \in G-H$. Let us have a look at $ax$. If this element would belong to $H$, then, since $x^{-1} \in H$, we would have $ax \cdot x^{-1}=a \in H$, which is not the case. So $ax \notin H$. Similarly, $ay, az \notin H$. Hence the set $K=\{a,ax,ay,az\}$ is disjoint from $H$. But now we have constructed $8$ elements. We conclude that amongst the elements of $K$, there must be two equals, say for example $a=ax$. This would give $x=e$, a contradiction. Similarly, the equality of any other pair of elements of $K$ would lead to a contradiction. 
A: 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.
