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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.

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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

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

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