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.


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.

  • 2
    $\begingroup$ Nit-pick --- Lagrange's Theorem implies that every element has order 5, 11, 55, or 1. $\endgroup$ – Gerry Myerson Jun 8 '13 at 4:31
  • $\begingroup$ Thank you Gerry and Michael. $\endgroup$ – Adam Saltz Jun 8 '13 at 4:45

Here is a simple elementary proof:

Is there any element that generate the whole group? If there is, you're done.

If not, all elements are order 5 or 11 (except the identity). Each element of order 11 generate a subgroup of order 11, and any two such subgroup generated this way are either identical, or share exactly only the identity. Similarly for order 5.

So if there are no order 5 element, then the group must have order $55=(11-1)k+1$ which is not possible. If there are no order 11 element, then the group must have order $55=(5-1)k+1$ which is also not possible.


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