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Let $G$ be a group such that the intersection of all its subgroups which are different from $\{e\}$ is a subgroup different from $\{e\}$ , then is it true that every element in $G$ has finite order ? I was trying to argue by contradiction , say $x \in G $ has infinite order , but actually couldn't go anywhere , Please help .

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Suppose $x \in G$ has infinite order. Then $\langle x \rangle \cong \mathbb{Z}$. The intersection of all the nontrivial subgroups of $\mathbb{Z}$ is $\{0\}$ ($0$ is the only number divisible by all the other numbers), thus the intersection of all the nontrivial subgroups of $G$ is reduced to $\{e\}$.

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  • $\begingroup$ (I've deleted my comment, so if you delete yours it will still be a hint, rather than a hint with an explanation below!) $\endgroup$ – user1729 Nov 3 '14 at 13:29
  • $\begingroup$ Wait - I'm confused again. We want to show that the implication "not all elements have finite order $\Rightarrow$ intersection is trivial" does not hold. But you have just given an example where it does hold. $\endgroup$ – user1729 Nov 3 '14 at 16:21
  • $\begingroup$ No, you want to show that that implication does hold. The statement you want to prove is: "If the intersection of all nontrivial subgroups is nontrivial, then every element of $G$ has finite order." The contrapositive of this statement, which is logically equivalent to the original statement, is: "If not every element of $G$ has finite order, then the intersection of all nontrivial subgroups is trivial." Proving that statement is equivalent to proving the original one. $\endgroup$ – Greg Martin Nov 3 '14 at 17:13
  • $\begingroup$ (PS: because of precisely this sort of confusion, I think it's good to recognize that proving the contrapositive is not the same as a proof by contradiction.) $\endgroup$ – Greg Martin Nov 3 '14 at 17:14
  • $\begingroup$ Oh, right, I understand. @GregMartin I thought it was meant as a counter-example (to the implication), rather than some sort of proof by contradiction. $\endgroup$ – user1729 Nov 3 '14 at 19:19

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