I Can a group be isomorphic to one of its quotients?

[WWGD]

Of course it must be an infinite group, otherwise |G/N|=|G|/|N| and then {e} is the only ( and trivial) solution. I understand there is a result that for every quotient Q:=G/N there is a subgroup H that is isomorphic to Q. Is that the case?

 

[Infrared]

How about an infinite product $\mathbb{Z}\times\mathbb{Z}\times...$ and quotient out by the first factor?

I understand there is a result that for every quotient Q:=G/N there is a subgroup H that is isomorphic to Q. Is that the case?

No, there is no subgroup of $\mathbb{Z}$ that is isomorphic to $\mathbb{Z}/2\mathbb{Z}.$

 

[WWGD]

How about an infinite product $\mathbb{Z}\times\mathbb{Z}\times...$ and quotient out by the first factor?No, there is no subgroup of $\mathbb{Z}$ that is isomorphic to $\mathbb{Z}/2\mathbb{Z}.$

Thanks. Sorry, I believe the result I quoted (may)apply to infinite groups.

 

[Infrared]

Thanks. Sorry, I believe the result I quoted (may)apply to infinite groups.

No, even if the quotient is infinite, it is still false. There is no subgroup of $\mathbb{Z}\times\mathbb{Z}$ that is isomorphic to $\left(\mathbb{Z}\times\mathbb{Z}\right)/\left(\{0\}\times 2\mathbb{Z}\right)\cong\mathbb{Z}\times\left(\mathbb{Z}/2\mathbb{Z}\right).$

 

[WWGD]

Re the lack of correspondence between subgroups and quotient groups, I though of another argument: just take a simple group. It will have non-trivial subgroups but no non-trivial quotient. Maybe simplest vase is $S_5$, the permutation group on 5 elements . It has the alternating subgroup, which cannot be a quotient by cardinality reasons.