This is something I'm supposed to be able to prove for an upcoming test, but I can't find anything to help me prove this in my notes or the chapter, which is on cosets and Lagrange's theorem. If all I start with is the group having no proper subsets, then that means any subset is the whole group. By Lagrange's theorem, that means the index is $1$, but that doesn't get me anywhere. Where do I start?
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2$\begingroup$ Do you know Cauchy's theorem? $\endgroup$– Alex WertheimCommented Nov 24, 2014 at 2:16
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3 Answers
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I'll give the proof of it without using Cauchy's theorem:
1. $G$ is cyclic If not, the generating set of $G$ must have at least two elements. Let $a$ and $b$ be elements which satisfy $a\notin \langle b\rangle$ and $b\notin \langle a\rangle$, then $\langle a\rangle$ and $\langle b\rangle$ are distinct proper subgroups of $G$, a contradiction.
2. $G$ has a prime order If $G$ has infinite order, then it has proper subgroup (consider $2\Bbb{Z}$ in $\Bbb{Z}$.) so $G$ is finite. If $|G|=rs$ with $r,s>1$ and $G=\langle a\rangle$, then $\langle a^s\rangle$ is a proper subgroup of $G$.
answered Nov 24, 2014 at 2:21
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$\begingroup$ This is really nice! Cauchy's theorem seems overkill when you put it this way. $\endgroup$ Commented Nov 24, 2014 at 2:24
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$\begingroup$ Ok, this makes sense to me and sounds good, but how would I justify the first sentence. If $G$ is not cyclic it must be generated by two elements. It makes sense but what theorem or property proves that it is true? $\endgroup$– chrisCommented Nov 24, 2014 at 2:41
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$\begingroup$ @chris We assume that $G$ is not cyclic, so $G$ is not generated by one element. $\endgroup$ Commented Nov 24, 2014 at 2:49
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$\begingroup$ Ho do you know that such an $a$ and such a $b$ exist? $\endgroup$ Commented Oct 20, 2016 at 18:07
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1$\begingroup$ @Jerrywu Excluding trivial cases, of course :) $\endgroup$ Commented Nov 19, 2021 at 19:54
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Here is a simple proof via Cauchy's theorem, proceeding by contrapositive.
Let $G$ be a group of composite order $n = pk$, $p$ prime. By Cauchy's theorem, $G$ has an element of order $p$, hence a cyclic proper subgroup of order $p$.
answered Nov 24, 2014 at 2:19
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1$\begingroup$ This may be a stupid question, but how can we be sure there is a cyclic proper subgroup of order $p$? Is it because the element with order $p$ generates a cyclic group containing just itself and the identity? $\endgroup$– chrisCommented Nov 24, 2014 at 2:29
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$\begingroup$ Any element $x\in G$ of order $\alpha$ generates a cyclic group of order $\alpha$. That is $\langle x \rangle = \{1, x, x^{2}, x^{3}, \ldots, x^{\alpha-1}\}$. $\endgroup$ Commented Nov 24, 2014 at 2:36
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I'm answering a 5-year old question because I had questions on this one. If G has no nontrivial subgroups then any of it's elements would generate it. If the order were composite, say $o(G)=nm \Rightarrow $ pick $a\neq e$ so that $a^{nm}=e=(\underbrace{a^{n}}_{b})^{m}=b^m\Rightarrow o(b)\leq m$.
Either $a^{n}=e$ or $b^{m}=e$. In either case, $G$ would have a nontrivial proper subgroup.
