# A group of order 66 has an element of order 33

If I wanted to show that a group of order 66 has an element of order 33, could I just say that it has an element of order 3 (by Cauchy's theorem, since 3 is a prime and 3|66), and similarly that there must be an element of order 11, and then multiply these to get an element of order 33? I'm pretty sure this is wrong, but if someone could help me out I would appreciate it. Thanks.

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This generally isn't true unless the two elements commute. –  Qiaochu Yuan Feb 28 '12 at 3:48
I figured this would be the case. Is there a way to repair my "proof"? –  johnq Feb 28 '12 at 3:50
Show that they commute! –  Mariano Suárez-Alvarez Feb 28 '12 at 3:52
It's not clear that they do necessarily commute. –  johnq Feb 28 '12 at 4:01
I did not say it was clear, I said prove it :) –  Mariano Suárez-Alvarez Feb 28 '12 at 4:02

Step 1: Show that any finite group order $4k+2$ has an index two subgroup. (Hint.)

Step 2: Show that any group of order $33$ is cyclic.

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Alternatively:

• show that the Sylow $11$-subgroup is normal.

• show that a cyclic group of order $11$ has no automorphisms of order $3$.

• pick an element of order $3$ in your group and conclude that it commutes with any element of order $11$.

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To compliment the other answers, I will address why commutativity is necessary. Let $G$ be your group $o(x)$ denote the order of $x\in G$. You are making use of the statement that $o(a)o(b)=o(ab)$, but this is not necessarily true. It holds when $o(a),o(b)$ are coprime and $ab=ba$ (an interesting consequence of this is that the partial sums $s_n$ of the harmonic series are never integers for $n>1$, which follows from Bertrand's postulate and $\mathbb Q/\mathbb Z$ being abelian), but there are nonabelian groups of order $66$, such as $S_6\oplus \mathbb Z_{11}$. If $ab=ba$ and $o(a),o(b)$ are coprime, then $(ab)^{o(a)o(b)}=a^{o(a)}b^{o(b)}=e\cdot e=e$, so $o(ab)\leq o(a)o(b)$. If $(ab)^n=e$ then $a^nb^n=(ab)^n=e$ so $(a^n)^{-1}=b^n$, hence $o(a^n)=o(b^n)$, and if $n<o(a)o(b)$ then since $o(a),o(b)$ coprime we have that one of $a^n,b^n\neq e$. But $o(b^n)=o(a^n)|o(a)$ since $(a^{n})^{o(a)}=(a^{o(a)})^n=e^n=e$, and similarly $o(a^n)=o(b^n)|o(b)$, so $o(a^n)=o(b^n)=1$. Thus $a^n=b^n=e$, so $n\geq o(a)o(b)$ so $o(ab)=o(a)o(b)$.