# If there exists an epimorphism from a group $G_1$ to $G_2$ that is not one-to-one, then can $G_1$ and $G_2$ be isomorphic? [duplicate]

I am trying to show that two groups are isomorphic only if a certain condition holds. I can show that a specific epimorphism between the two groups is one-to-one only if this condition holds, but it isn't obvious that for infinite groups that this should mean that there can't exist an isomorphism between the two groups.

If there is a non-one-to-one epimorphism from a group $$G_1$$ to a group $$G_2$$, does this mean that all epimorphisms from $$G_1$$ to $$G_2$$ are not one-to-one?

## marked as duplicate by Paul Plummer, user10354138, Lord Shark the Unknown, Derek Holt group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 22 '18 at 7:55

• No. The free abelian group $\mathbb Z^\infty$ has a quotient (e.g. quotienting out the first component) isomorphic to itself. – Cave Johnson Nov 22 '18 at 1:35
No, that would be too much to ask. For instance, consider the map $$g:(\Bbb Z[X],+)\to (\Bbb Z[X],+)\\ g(p)=\frac{p-p(0)}{X}$$
Then, $$\ker g=\Bbb Z$$, the map $$p\mapsto Xp$$ is a section of $$g$$ and $$id:\Bbb Z[X]\to \Bbb Z[X]$$ is an isomorphism.