# If $G$ is isomorphic to all non-trivial cyclic subgroups, prove that $G\cong \mathbb{Z}$ or $G\cong \mathbb{Z}_p$

Assuming $G$ is isomorphic to all its non-trivial subgroups, prove that $G$ is isomoprhic to either $\mathbb Z$ or $\mathbb Z_p$ for $p$ prime.

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Looks like a homework. Then please add a tag. –  Ilya Oct 20 '11 at 16:07
You will go a long way by rephrasing your question to be like "Please help me prove..." Also it would help to know what work you've done on this problem already. –  Unreasonable Sin Oct 20 '11 at 16:15
I apologize for the pedantry, but (a) the title question is different from the question in the body, and (b) the assertion in the body is false, as $G$ might be the group with one element. –  user83827 Oct 20 '11 at 17:11

There are two a priori possibilities:

(1) It could be that $G$ has no nontrivial subgroups; then the condition "$G$ is isomorphic to each one of its nontrivial subgroups" is true by vacuity. This can only happen if $G$ itself is trivial (if $G\neq\{1\}$, let $x\in G$, $x\neq 1$; then $\langle x\rangle$ is a nontrivial subgroup of $G$). So we could have $G=\{1\}$, not contemplated in the original question.

(2) It could be that $G\neq\{1\}$. Then $G$ must be cyclic: if $x\in G$, $x\neq 1$, then $\langle x\rangle$ is a nontrivial subgroup of $G$, hence $G\cong\langle x\rangle$ is cyclic. Thus, either $G\cong\mathbb{Z}$, or $G\cong C_n$ for some $n\gt 1$.

If $G\cong\mathbb{Z}$, then $G$ is isomorphic to any nontrivial subgroup.

If $G\cong C_n$, and $d|n$, then $G$ has a subgroup of order $d$ which is cyclic. If $d\gt 1$, then $G\cong C_d\cong C_n$, so $d=n$. Thus, the only divisors of $n$ are $1$ and $n$, hence $n$ is prime. We change its name to $p$, because that is the only way in which the original statement will be true, and we conclude that $G\cong C_p$, as desired.

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why is it in (2) that G must be cyclic. I first want to show that G is cyclic if G is isomorphic to all its non trivial cyclic subgroups, and then i must show that G is isomorphic to either Z or Zp –  Eidbanger Oct 21 '11 at 16:04
@Eidbanger: I explain why $G$ must be cyclic in the next sentence: pick any $x\in G$, $x\neq 1$. Then $\langle x\rangle$ is a nontrivial subgroup of $G$, and is certainly cyclic. Since $G$ is isomorphic to every one of its nontrivial subgroup, then $G\cong \langle x\rangle$, and therefore $G$ is cyclic (because it is isomorphic to a cyclic group). And every cyclic group is either isomorphic to $\mathbb{Z}$ (if it has infinite order), or to a finite cyclic group of order $n$ for some $n$. –  Arturo Magidin Oct 21 '11 at 17:31
By your hypothesis i would assume $|G|>1$ otherwise there are no non trivial subgroups and everything loses interest.
However, suppose $|G|$ is finite. Then there exists a prime $p$ such that $p| |G|$. Hence use Cauchy to find an element $g\neq e$ such that $|g|=p.$ Then consider $\langle g\rangle\cong C_p$ where $C_p$ is a cyclic group of order $p$. But then $G\cong C_p\cong \mathbb Z_p$.
On the other hand, suppose $|G|=+\infty.$ Then there must exist an element of infinite order, for otherwise we should consider the cyclic group generated by an element, and if it were finite, it would be impossible for this non trivial subgroup to be isomorphic to $G$. But then consider this $g\in G$ of infinite order. Then $G=\langle g\rangle$, and we know that every infinite Cyclic group is isomorphic to $\mathbb Z$, hence $G\cong \mathbb Z$.
EDIT In answering to the op comment, i would like to point out that there is no really need to appeal to Cauchy theorem if $|G|$ is finite. Indeed, since $G$ is finite, pick an element in $|G|$, say $j$, different from the identity element, and compute its order. Then this order is $n$, with $n\mid |G|$ by lagrange theorem. We are not done yet since we are looking for a prime $p$, but to do so note that if $n$ is decomposable as $n=p\cdot m$, then $g=j^m\in G$ and $|j^m|=p$. Since $G$ is isomorphic to all its trivial subgroups, then $G\cong \langle g \rangle=\langle j^m\rangle\cong C_p$, the cyclic group of order $p$, and then conclude as above.