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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 $e$. Prove that every element of $G$ has finite order.

Can i get some hints to get started?

Thanks.

EDIT

As per suggestion of user Slade, i have proceeded according to this

If $g$ has infinite order, then the cyclic group that $g$ generates will have elements of the form $\{g,g^{2},g^{3},......)$.

The claim is that no two elements are same, i.e. there are no repetitions. To show this let us assume that there are repetitions, say $a^{i}=a^{j}$, for some $i,j$. Then $a^{i-j}=e$ and this contradiction of $a$ having infinite order.

What should i do next?

JKnecht
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Taylor Ted
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4 Answers4

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Suppose that $G$ has an element $g$ of infinite order. What are the subgroups of the cyclic group that $g$ generates?

Andrew Dudzik
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  • Coudnot quite follow you – Taylor Ted Mar 24 '15 at 19:40
  • @K.Dutta Which part? – Andrew Dudzik Mar 25 '15 at 01:04
  • If g has infinite order ,then cyclic group that g generates will have elements of form {g,$g^{2}$,$g^{3}$,......) .Claim is that no two elements are same .ie there are no repetitions .To show this let us assume that there are repetitions say $a^{i}=a^{j}$ for some i,j .Then $a^{i-j}=e$ this contradicts of us having infinite order of a .What should i do next – Taylor Ted Mar 25 '15 at 03:14
  • @K.Dutta Don't forget the identity element, and don't forget inverses. Next, you should find all nontrivial subgroups of an infinite cyclic group, and calculate their intersection. – Andrew Dudzik Mar 25 '15 at 04:41
  • Does G = it have two subgroups generated by g and $g^{-1}$ – Taylor Ted Mar 25 '15 at 04:46
  • @K.Dutta $g$ and $g^{-1}$ generate the same subgroup, namely all of $\langle g\rangle$. But there are many others as well. It may help to realize that an infinite cyclic group can be identified with $(\mathbb{Z},+)$. – Andrew Dudzik Mar 25 '15 at 04:52
  • I coudnot see this .please provide detaisl on how to to do next – Taylor Ted Mar 25 '15 at 04:54
  • @K.Dutta Every subgroup of a cyclic group is cyclic. What group is generated by $g^2$? By $g^3$? For this problem, you don't need to find them all, you just need to verify that the intersection of the subgroups different from ${e}$ equals ${e}$ itself. – Andrew Dudzik Mar 25 '15 at 05:01
  • subgroup generated by $ g^{2}= $ {$g^{2}$ , $g^{4}$ , $g^{6}$ ..) and by $g^{3}$ = {$g^{3}$ , $g^{6}$ , $g^{9}$..next ?? – Taylor Ted Mar 25 '15 at 05:07
  • @K.Dutta You're close, but keep in mind that a subgroup needs to contain the identity element and inverses. – Andrew Dudzik Mar 25 '15 at 05:12
  • e is contained in both of them .Now if i prove that no to elements are same in both of these subgroups than identity itself ,will that be way – Taylor Ted Mar 25 '15 at 05:14
  • @K.Dutta These subgroups both contain $g^6$ (and also $g^{-6}$, $g^{12}$...). You need to look at more subgroups. – Andrew Dudzik Mar 25 '15 at 05:46
  • please tell answer – Taylor Ted Mar 25 '15 at 05:48
  • @K.Dutta Please describe all subgroups of $(\mathbb{Z},+)$. You will find this in the earliest chapters of any textbook on the subject. – Andrew Dudzik Mar 25 '15 at 05:51
  • Sungroups are nZ – Taylor Ted Mar 25 '15 at 05:53
  • @K.Dutta And which integers are a multiple of $n$, for every $n$? – Andrew Dudzik Mar 25 '15 at 05:56
  • means? i cannot understand .can you rephrase it – Taylor Ted Mar 25 '15 at 05:58
  • @K.Dutta You are trying to compute $\bigcap_n n\mathbb{Z}$. – Andrew Dudzik Mar 25 '15 at 05:59
  • their intersection will be lcm of all multiples of n – Taylor Ted Mar 25 '15 at 06:02
  • @K.Dutta And what is the least common multiple of all the integers $1,2,3,\cdots$? – Andrew Dudzik Mar 25 '15 at 06:04
  • it will be 1 or product of all these – Taylor Ted Mar 25 '15 at 06:05
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    @K.Dutta Incorrect; $1$ is not a multiple of $2$, for example. There is only one integer that is a multiple of all other integers. – Andrew Dudzik Mar 25 '15 at 06:06
  • please enlighten me – Taylor Ted Mar 25 '15 at 06:08
  • @K.Dutta Please spend more than two minutes thinking about it, and ask again if you are still stuck after some time. – Andrew Dudzik Mar 25 '15 at 06:11
  • Guys. I approve of this method of teaching. In fact I more than approve of it! As a moderator I have a small request: When you are done with this exchange, then could you please delete those comments that are no longer needed. For the benefit of future readers the main points may (at that point) be edited into Slade's answer. This is just because the length of the chain of comments triggers a flag that summons a moderator to the scene. – Jyrki Lahtonen Mar 25 '15 at 06:12
  • One more thing. Because I have littered the site with similar comment exchanges myself, I cannot with a clear conscience enforce this SE-norm :-). So, Slade, strictly your own call whether you want to do that! I am clearing the flag anyway! – Jyrki Lahtonen Mar 25 '15 at 06:14
  • @JyrkiLahtonen I'll tidy up shortly. Thanks for the encouragement on both my tidiness and my pedagogy. – Andrew Dudzik Mar 25 '15 at 06:16
  • @Slade please tell how to proceed with this – Taylor Ted Mar 25 '15 at 06:37
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Observe that in infinite cyclic group that intersection is just the identity. Conclude that your group has not a cyclic infinite subgroup. This evidently equivalent to being every element of finite order

Hhhh
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Hint by absurd, take an element $a$ without a finite order and consider $I$ is the intersection of subgroups of $<a>$ different from $\{e\}$. prove that : $$I=\{e\} $$ (if $x=a^m \in I$ then $x\in <a^{m+1}>\cap <a^{m+1}>\cdots$)

Elaqqad
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A direct approach would be to take some $g\neq e$ which is contained in all nontrivial subgroups (which exists by hypothesis), and note that for any $x\neq e$, the set $\{e,x,x^2,x^3,\ldots\}$ forms a nontrivial subgroup and hence contains $g$. Thus, for any $x\neq e$ we have $x^n=g$ for some $n$. We can first use this to show that, letting $x=g^2$ we have either that $g^{2}=e$ or $g^{2n}=g$ for some $n$ - either of which means $g$ has finite order - that is, it satisfies $g^m=e$ for some $m>0$. However, since any $x$ other than the identity has a finite $n$ such that $x^n=g$ we have $x^{nm}=g^m=e$, hence every element has finite order.

Milo Brandt
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