Abelian group G such that the infinite-order elements form a subgroup with the identity. If {g∈G∣g=e or g has infinite order} is a subgroup of G, what can we say about the order of the elements of G?

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Abelian group G such that the infinite-order elements form a subgroup with the identity.  If {g∈G∣g=e or g has infinite order} is a subgroup of G, what can we say about the order of the elements of G?

Jimmy Macias · Accepted Answer

Step 1  Suppose otherwise. Let a have infinite order and b be nontrivial with finite order. Then ab has infinite order, since a has infinite order, because otherwise (ab)n=e implies b−n−an, a contradiction. But observe that  b=eb=aa−1=(ab)a−1  is an element of the candidate subroup H (as a−1,ab∈H), a contradiction.

Corgnatiui · Accepted Answer

Step 1 Let H={g∈G∣g=e or g has infinite order} G contains all elements of finite order or all elements (except identity) of infinite order. And other possibility is impossible. Because if, ∃a,b∈G such that |a|=n(&amp;gt;1) and |b|=∞ then it will contradict that H is a subgroup of G. It violates closure property. b∈H⇒b−1∈H And ab∈H But, abb−1=a⧸∈H |a|=m⇒am=e If ab⧸∈H, then, |ab|=finite=n (say) Then, (ab)mn=e Rig↔ow(am)nbmn=e Rig↔owbmn=e Hence, |b| divides mn Contradict, |b|=infinite

Abelian group G such that the infinite-order elements form a

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