# Every Group of Order 9 has an element of order 3

Prove that every group of order 9 has an element of order 3.

So my attempt at this proof was:

Pf. Let $G$ be a group of order 9. Then $\exists H$ such that $H$ is a subgroup of $G$ and $|H|$ $\vert$ $|G|$. Thus $|H|=$ 1,3, or 9 by Lagrange's Theorem.

I know this proof is not complete, but I am unsure on how to proceed from here.

You're almost there… each element of G has an order which divides 9 so the order of each $g\in G$ is 1,3 or 9.
case 1: $ord(g) = 1$, then it's the neutral because there can only be 1 of it, the other 8. need to be case 2 or 3
case 2: $ord(g) = 3$, you are done
case 3: $ord(g) = 9$, consider $g^3$ which has order 3