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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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:
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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 $o(ab)=o(a)o(b)$. |
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