Finite subgroup of the multiplicative group of a field is cyclic

Asked Feb 05 '19 at 06:56

Active Feb 05 '19 at 06:56

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Dummit and Foote's Abstract Algebra contains a proof that a finite subgroup of the multiplicative group of a field, $F$, is cyclic. However the seems unnecessarily complex to me. Certainly their version works (pictured below), but I'm curious why they don't simply take the approach of adding $(1,...,1)\in(\mathbb{Z}/n_1\mathbb{Z}\times\cdots\mathbb{Z}/n_k\mathbb{Z})$ to itself $n_1\cdot n_2\cdots n_k$ times to show that the group necessarily has a generator. Does that approach somehow fail?

Dummit & Foote's proof:

asked Feb 05 '19 at 06:56

JFox

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How would that show that the group has a generator? – Eric Wofsey Feb 05 '19 at 06:58
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If $k>1$, then that group isn't cyclic, so one has to prove $k=1$. – Angina Seng Feb 05 '19 at 07:13
Why is the group not cyclic if k>1? – JFox Feb 05 '19 at 07:14
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Keep in mind that, in an additive group, if $nx=0$, that does not show that $x$ has order $n$, only that it has order dividing $n$. – verret Feb 05 '19 at 07:15
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@JFox It might help you to consider an explicit example. Take $Z/2Z \times Z/2Z$, and see what elements you get if you add $(1,1)$ repeatedly to itself. – verret Feb 05 '19 at 07:15
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Ah I see, I think I was imagining $n_1,...,n_k$ as coprime but yeah that doesn't make sense. Thanks for the help. – JFox Feb 05 '19 at 07:20
Yes, in fact, it's the complete opposite, in this situation, we have $n_k\mid\cdots\mid n_1$. – verret Feb 05 '19 at 08:22

Finite subgroup of the multiplicative group of a field is cyclic

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