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

• I
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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?

## Answers and Replies

Science Advisor
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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}.##

heff001, mathwonk, graphking and 2 others
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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.

Science Advisor
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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
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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.