New Criteria of Isomorphism + Abelian Group

Question

The following questions were suggested by my friend, while we were studying fundamental group theory. We had no exact ideas of the way to approach the problems.

Questions

(1) Let $G$ and $H$ be groups such that $|H|=|G|$. If we can make a bijection $\phi:G\to H$ such that for $\forall a\in G$, $ord(a)=ord(\phi(a))$, then does $G\cong H$?

(2) Let $G$ be a group. If all of the elements in $G$ except $e$ have the same order, then is $G$ an abelian group?

Both suggestions were not easy to find the counter-examples. So, we wanted to prove these statements instead, but again too difficult. We also failed to find these types of questions on the internet. Is there theories or propositions related to the problems? Thanks.

For $(1)$ see here. For $(2) $ see here , the Heisenberg group over $\Bbb F_p$. So both questions are duplicates. — Dietrich Burde, Jun 16 '20 at 11:07
Concerning $2$ : For a counterexample see here : https://groupprops.subwiki.org/wiki/Prime-cube_order_group:U(3,3) — Peter, Jun 16 '20 at 11:08
Note that all counterexamples to (2) must be groups of prime power order, by Cauchy's theorem. — Stahl, Jun 16 '20 at 11:09
@freakish Good point! For some reason I had been assuming we were necessarily working with finite groups. — Stahl, Jun 16 '20 at 11:11

freakish · Accepted Answer · 2020-06-16T11:22:10.653

(1) Let $G$ and $H$ be groups such that $|H|=|G|$. If we can make a bijection $\phi:G\to H$ such that for $\forall a\in G$, $ord(a)=ord(\phi(a))$, then does $G\cong H$?

Yes, if both are finite abelian. This follows from the classification of finite abelian groups, see here.

No, if both are infinite, even abelian. The simple counterexample is $\mathbb{Z}$ and $\mathbb{Z}^2$.

No, if one of them is non-abelian, even when both finite: the Heisenberg group over $\mathbb{Z}_p$ and the corresponding $(\mathbb{Z}_p)^n$.

For a deeper discussion on the subject read this: Is a finite group uniquely determined by the orders of its elements?

(2) Let $G$ be a group. If all of the elements in $G$ except $e$ have the same order, then does $G$ an abelian group?

No. Neither in finite nor in infinite case (with finite order). For finite case we have the already mentioned Heisenberg group, for infinite case the Tarski monster group.

@Peter Ups, for some reason I thought we talk about abelian groups in (1). Fixed. — freakish, Jun 16 '20 at 11:17

New Criteria of Isomorphism + Abelian Group

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