If $|G| = 55$, must it have an element of order $5$ and/or $11$?
I'm not quite sure how to determine this. I know it could be possible by Lagrange's Theorem, but I'm stuck otherwise. Any help would be appreciated.
Edit: I haven't learned material about Cauchy's or Sylow Theorems yet. So I'm trying to prove this by very elementary facts.
Best Answer
Yes. Lagrange's theorem implies that every element has order 5, 11, or 55.
If there is an element of order $55$, call it $g$. Then $g^{11}$ is an element of order $5$ and $g^5$ is an element of order $11$.
So let's suppose that there are no elements of order $55$. If there are elements of order $5$ and $11$, then we are done. So suppose there are only elements of order $11$. Such an element generates a cyclic subgroup of $10$ non-identity elements. Moreover, any other element in the subgroup generates the same subgroup. Now pick another element in $G$ not in that cyclic subgroup. Each time we do this, we get $10$ more non-identity elements. But then the group must have $11, 21, 31, 41, 51, 61,\ldots$ elements. In particular, such a group cannot have $55$ elements.
A similar argument shows that a group with elements only of order $5$ cannot have $55$ elements.
Unwinding all of this, we see that we reach a contradiction unless there exist elements of order $5$ and $11$ in the group.