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

  • $\begingroup$ Slightly embarrassed because it was so obvious. Thank you! $\endgroup$ – El Spiffy Nov 10 '16 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.