Every element of a group has finite order

Asked Apr 25 '18 at 14:28

Active Apr 25 '18 at 14:52

Viewed 66 times

Suppose that $G$ is a group, and that the intersection over the non-trivial subgroups of $G$ is non-trivial, i.e., that $$\bigcap_{H<G\atop H\ne \{e\}} H\neq\{e\}.$$ I need to demonstrate that every element of $G$ has finite order.

I tried showing that if the intersection is trivial, an element of infinite order would be there, but couldn't get anywhere.

edited Apr 25 '18 at 14:52

peter a g

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asked Apr 25 '18 at 14:28

Daniel Moraes

So your assumption is $$\bigcap_{H\leq G} H\neq{e}?$$ – Dave Apr 25 '18 at 14:32
@Dave yes, Sorry not Very used to LaTeX i wrote It a litlle confuse – Daniel Moraes Apr 25 '18 at 14:34
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This can never happen though. The trivial group is a subgroup of any group $G$, and since it is the "smallest group", we always have $$\bigcap_{H\leq G} H={e}$$for any $G$. – Dave Apr 25 '18 at 14:35
@Dave yes, but it is saying there is someone Else there, not only $e$ – Daniel Moraes Apr 25 '18 at 14:37
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But this can never happen. If you're taking the intersection over all subgroups $H$ of a group $G$, then the identity is the only common element in the $H$'s. – Dave Apr 25 '18 at 14:38
@user528821 - do you mean to take the intersection over NON-TRIVIAL $H$ - i.e., with $H\ne {e}$? (going along with Dave's point.) – peter a g Apr 25 '18 at 14:40
@Dave Sorry, forgot an hipotesys, the subgroup of only $e$ is not in the intersection – Daniel Moraes Apr 25 '18 at 14:40
@peterag exactly, Sorry guys, It was not explicit in the exercise, Saw It now but still stuck – Daniel Moraes Apr 25 '18 at 14:41
@Dave thanks, should i delete the question? – Daniel Moraes Apr 25 '18 at 14:44

Every element of a group has finite order

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