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In the link below, someone has made a comment that an abelian group of order $n$ with $n=p_1...p_n$, where $p_1...p_n$ are primes, is cyclic. Can someone please explain why this is the case?

Thanks in advance!

An abelian group of order $n$ with $n=p_1...p_n$, where $p_1...p_n$ are primes, is cyclic

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Frank Booth
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  • You can use the fundamental theorem of finite abelian groups...but that is kind of a sledgehammer. To avoid the sledgehammer, you can generalize the proof in that post. Use Cauchy's theorem to produce an element of order $n$. – John Martin Apr 12 '16 at 18:34
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    The condition $n=p_1p_2\cdots p_n$ is impossible since $p_1p_2\cdots p_n>n$ for all $n\ge 1$. – Dietrich Burde Apr 12 '16 at 18:36
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    I think the $n's$ are supposed to be different? That's how I was reading it anyway. – John Martin Apr 12 '16 at 18:37
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    @JohnMartin Yes, but better to ask what exactly the question is. – Dietrich Burde Apr 12 '16 at 18:38
  • Use Sylow's first theorem to decompose the abelian group as a direct sum of its sylow-p subgroups and use the fact that $\mathbb{Z}{m} \oplus \mathbb{Z}{n} \cong \mathbb{Z}_{mn}$ whenever $m$ and $n$ are relatively prime, along with induction. – iron feliks Apr 12 '16 at 18:38

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The hard part is the existence of elements of order $p_i$ for each $i$. This follows from Cauchy's theorem, which has a simple induction proof for finite abelian groups.

Once you have those elements, the result follows by repeated application of this general fact:

In an abelian group, if $a$ has order $m$ and $b$ has order $n$, and $m$ and $n$ are coprime, then $ab$ has order $mn$.

Indeed, if $(ab)^k=1$ then $a^k=b^{-k} \in \langle a \rangle \cap \langle b \rangle = 1$ because $m$ and $n$ are coprime. Therefore, $a^k=b^k=1$ and so $k$ is a multiple of $lcm(m,n)=mn$.

lhf
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