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Suppose $G\neq\{e\}$ is a group whose only subgroups are $\{e\}$ and G.

Prove that $G$ is finite and $|G|$ is prime.

In regards to the prime part, I believe that we can simply apply Lagrange's theorem, but first I must prove that G is not infinite. Generally one would start something like this by assuming that G is infinite and try to find a contradiction. Although I'm not exactly sure where to start with this. Any help would be greatly appreciated.

DrkVenom
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  • Lagrange's theorem doesn't quite work. What if $|G|=p^2$, and your subgroup has order $p$? Do you know Cauchy's theorem? – Joshua Pepper Mar 06 '14 at 17:35
  • @JoshuaPepper No, unfortunately I'm not familiar with Cauchy's theorem. But how would Lagrange's not work once we prove it is finite. Since the order of a subgroup must divide the group, if the only subgroups are itself and the trivial subgroup, then the order must be prime, since only p and 1 divide p. – DrkVenom Mar 06 '14 at 17:39
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    Your argument shows that $|G|$ can have no proper subgroup, assuming $|G|=p$ is prime. You want to show the converse. – Joshua Pepper Mar 06 '14 at 17:41
  • Hint: what would be an easy way of constructing/identifying a subgroup which has a chance of being non-trivial? – Mark Bennet Mar 06 '14 at 17:43
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    @DrkVenom : Lagrange works in one direction only. It gives info on the number of elements in a subgroup, but is not good to show existence of subgroups. – Andrea Mori Mar 06 '14 at 17:47

2 Answers2

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Start with any nonidentity element $g \in G$. We see that the cyclic subgroup $\langle g \rangle$ generated by $g$ is in fact equal to $G$, so $G$ is cyclic. If $G$ were infinite, we would reach a contradiction.

Knowing that $G$ is finite, we have $G \cong \mathbb{Z}/n\mathbb{Z}$ for some integer $n$. What if $n$ were not prime?

Nick
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Nick's answer is ideal. Here's another, less ideal, solution. If $G$ is infinite and cyclic, say $G=\langle x \rangle$ then $\langle x^2 \rangle\subsetneq G$. Otherwise there is some element for which $\langle x \rangle\subsetneq G$. So $G$ is finite.

If $G$ has order $p^r m$ where $p\not|\ m$ then it has a subgroup of order $p^r$ by Sylow's Theorems. So $G$ must have a subgroup having prime-power order, meaning $G$ must have prime-power order.

In order for $G$ to not have nontrivial subgroups, this question shows that power must be 1.

Community
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David P
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