A group with $|G| = 33$ must contain an element of order $11$. Prove or disprove.

This is inspired by another MSE question. So we know that there must be an element with order 3.

I tried using Lagrange's theorem and similar, but no luck so far.

EDIT: (context clarification) I would prefer a solution that would not rely directly on Cauchy's theorem - some solution similar to answers to another question I linked to. Let's say this was a problem from a math class, and Cauchy's theorem had not been mentioned.

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    $\begingroup$ Have you studied Cauchy's Theorem (of group theory, of course), or Sylow theorems? $\endgroup$ – Timbuc Oct 30 '14 at 5:06
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    $\begingroup$ use cauchy's theorem for prime p=11 $\endgroup$ – Learnmore Oct 30 '14 at 5:07
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    $\begingroup$ That's a very strange edit... If you do know Cauchy's theorem, what prevents you from directly applying it here? Hint: 11 is prime. $\endgroup$ – Najib Idrissi Oct 31 '14 at 13:23
  • $\begingroup$ I would prefer solution that would not rely on Cauchy's theorem. @Nadib $\endgroup$ – VividD Oct 31 '14 at 13:25
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    $\begingroup$ If you don't want to rely on Cauchy's theorem, you can try to mimic the proof (without ever mentioning the name). $\endgroup$ – jdoicj Oct 31 '14 at 13:35

I'm not sure if this meets your goals. Relying heavily on Lagrange and basic bits about conjugacy classes.

Assume that a group $G$ with $33$ elements, none of order eleven, exists. If the group had an element $x$ of order $33$, then $x^3$ would be of order $11$, so no such $x$ exists. By Lagrange, all the non-identity element of $G$ must have order three.

Assume first that $G$ has a non-trivial center. Let $a\neq 1$ be a central element. It generates a normal subgroup $N$ of order three. The quotient group $G/N$ is then of order $11$, and thus cyclic. If $xN\in G/N$ is of order $11$, then the order of $x$ in $G$ must be divisible by $11$ - a case we already dealt with.

So we know that $G$ must have a trivial center. Let $a$ be any element of order three. It generates a subgroup $H$ of order three that centralizes $a$. Because $a$ is not in $Z(G)$, we know that the centralizer $C_G(a)=\{x\in G\mid xa=ax\}$ is a proper subgroup of $G$ containing $H$. By Lagrange there cannot be any subgroups properly between $H$ and $G$, so this implies that $C_G(a)=H$.

This implies that $a$ has exactly $11=[G:H]$ conjugates. So all the non-identity elements are partitioned into conjugacy classes of size eleven. But there are 32 of them, so this cannot be. QED

Of course, the class equation is lurking in there, so this comes very close to using Cauchy.

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    $\begingroup$ Actually, I think that this question is a good illustration that if you don't assume Cauchy or Sylow directly, you have to use the class equation in some guise to get an element of order $11$. $\endgroup$ – Geoff Robinson Jan 13 '15 at 16:52

Without using Cauchy's theorem one way to approach this problem is with the more powerful Sylow's Theorem. If $\vert G \vert = 33$ then we know that $G$ has at least one Sylow $p$-subgroup for every prime $p$ that divides $G$. Specifically looking at $p = 11$ there is a subgroup $K$ of $G$ with $\vert K \vert = 11$ (the maximal size for a subgroup of order $11^n$ in $G$). Now the only group of order $11$ is $\mathbb{Z}_{11}$ so we know that $K \cong \mathbb{Z}_{11}$. therefore there are elements of $K$ (all the non identity elements) that have order $11$.

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    $\begingroup$ I suspect that the proof of existence of Sylow subgroups uses Cauchy's theorem. $\endgroup$ – Daniel Fischer Oct 31 '14 at 16:18
  • $\begingroup$ @DanielFischer in John Rose's "A Course on Finite Groups" there is a proof of Sylow's Theorem without using Cauchy's theorem. He specifically notes that this is the case and so we can use the theorem to (re)prove Cauchy's Theorem (in terms of the flow of the book the theorem had already been proved so that is why I add re). $\endgroup$ – user171177 Oct 31 '14 at 16:21
  • $\begingroup$ @DanielFischer you are right that some proofs probably do use Cauchy's Theorem but it isn't needed. $\endgroup$ – user171177 Oct 31 '14 at 16:22
  • $\begingroup$ Ah, cool. Is Rose's proof short enough to give a sketch here? $\endgroup$ – Daniel Fischer Oct 31 '14 at 16:24
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    $\begingroup$ Now you mention it, I have seen that proof. Had completely forgotten about it, however. $\endgroup$ – Daniel Fischer Oct 31 '14 at 16:35

Sylow's theorems say since $33=3\cdot 11$, then there exists a Sylow $p$-Subgroup of order 11 and that the number of these groups is congruent to $1\pmod {11}$ and divides 33. This means that the group of order 11 is unique. Since this subgroup is of prime order, it is generated by a single element of order 11 in $G$.


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