If a group $G$ has only finitely many subgroups, then show that $G$ is a finite group. I have no idea on how to start this question. Can anyone guide me?
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4What happens if $G$ be assumed to be infinite? – Mikasa Mar 06 '13 at 16:48
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Duplicate of http://math.stackexchange.com/q/22996/264 – Zev Chonoles Mar 06 '13 at 16:52
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See this: http://math.stackexchange.com/questions/285726/on-subgroups-of-a-group – Mar 06 '13 at 16:55
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Suppose $G$ were infinite. If $G$ contains an element of infinite order, then _.
Otherwise, every element of $G$ has finite order, and if $x_1, x_2, \ldots, x_n$ are any finite set of elements of $G$, then the subgroups $\langle x_i \rangle$ cover only finitely many elements of $G$. Therefore __.
Ted
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I think the negative form of that is easier to prove: "If a group $G$ is infinite, then it has infinitely many subgroups".
What I would then do is take a member $a\in G$ such that $\left< a \right>$ is infinite, then prove there are an infinite number of subgroups just there, and say that if no such $a$ exists, there must be infinitely many subgroups of the type $\left< a \right>$.
Alfonso Fernandez
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Hint: The infinite cyclic group $\mathbb Z$ has infinitely many subgroups ($n\mathbb Z$) so every element of $G$ must have finite order. Use this to show that there is a finite collection $g_1, \ldots, g_n \in G$ of elements such that every element of $G$ is of the form $g_i^j$ for some choice of $i$ and $j$.
Jim
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