Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theorems I'm showing that any group of order $63$ has an element of order $3$, and can only use Lagrange's theorem not Cauchy's or Sylow's. I got it reduced to a case of having $62$ elements of order $7$ but I'm stuck now.

So let $|G|=63$, and consider an element $a \in G$, $a$ not equal to the identity. I'm allowed to use the theorem that if $|a|=n$, and $k$ divides $n$, then $|a^{n/k}|=k$.  Also, I know $G$ can only have elements of orders $1,3,7,9,21,63$.  If $|a|=63$, then $|a^{21}|=3$.  If $|a|=21$, then $|a^7|=3$.  If $|a|=9$, then $|a^3|=3$.  If $|a|=3$, then we are done. So $G$ must contain an element of order $3$ unless $G$ contains only the identity, and $62$ elements of order $7$. Now how do rule out this?
Hint: Let $G$ be your group of order $63$, and suppose every non-identity element has order $7$. Show that you can put an equivalence relation on $G\setminus\{e\}$, defined by $a\sim b$ when $a$ and $b$ generate the same subgroup of $G$. What are the sizes of the equivalence classes? Can they partition $G\setminus\{e\}$?
|G|=63 By Lagrange theorem if a belongs to G and a is non identity element
Possible orders of  a= 1,3,6,7,9,21,63
If | a | =63 , order of a^21= 3
If  | a |= 21, order of a^7 =3
If | a | =9, order of a^3= 3
If | a | =6 ,  order of a^2=  3
If a has order 63, 21, 9, 6 and of course 3 then there is an element of order 3 in G
So if a does not have order 3 all non identity(62 ) elments of G must have order 7
But we know for |G |= 63 no of elements of order 7= 6k ( a multiple of 6)
But 62 is not a multiple of 6 hence there must be a non identity element of G whose order is 3.