# 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?

• Can you show your work? – Pedro Tamaroff Feb 18 '14 at 5:37

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 $G\setminus\{e\}$?
• The identity element is not involved because we're working in the set $G\setminus\{e\}$ (also it generates the trivial subgroup, not a subgroup of order $7$). – Zev Chonoles Feb 18 '14 at 5:57
• Take a subgroup of $G$ of order $7$. How many elements generate it? – Zev Chonoles Feb 18 '14 at 6:00