If a group and its quotient group are isomorphic, the normal subgroup only has identity element?

Question

Let $K$ is a normal subgroup of a group $G$,

if $G \cong G/K$, $K=\{ e \}$?

Firstly, the converse is true.

If $G$ is a finite group, it is obviously true.

But if $G$ is an infinite group, I've no idea how to prove it or to come up with any counterexample.

If $G$ is finie yes, but otherwise I don't think it's true... — user330587, Jun 26 '18 at 16:38
Hint: $z \mapsto z^2$ as a morphism $\Bbb C^{\times} \to \Bbb C^{\times}$ — Watson, Jun 26 '18 at 16:49

David C. Ullrich · Accepted Answer · 2018-06-26T17:01:41.853

2

Counterexample: Say $G$ is the group of sequences of integers, with addition. Let $K$ be the subgroup consisting of sequences $(n_1,0,0,\dots)$.

In slightly more detail: An element of $G/K$ is a coset $K+(n_1,n_2,\dots)$. Define $\phi:G\to G/K$ by $\phi((n_1,n_2,\dots))=K+(0,n_1,n_2,\dots)$. Then you can show $\phi$ is an isomorphism. (It's clear that $\phi(n+m)=\phi(n)+\phi(m)$ and that $\phi$ is surjective; it's not hard to see that if $\phi(n)=0$, that is, $\phi(n)=K+(0,0,\dots)$, then $n=0$; hence $\phi$ is injective.)

Oops, maybe it's not clear that $\phi$ is surjective. Regarding that, note that $K+(n_0,n_1,n_2,\dots) = K+(0,n_1,n_2,\dots)$.

edited Jun 26 '18 at 17:01

answered Jun 26 '18 at 16:39

David C. Ullrich

89,985

Thx. Then will the conjecture be true when $|G| \le \aleph_0$? – namasikanam Jun 26 '18 at 17:17
1

@TA123 No. Same argument, with $G$ the group of all sequences of integers with only finitely many non-zero entries. – David C. Ullrich Jun 26 '18 at 17:18

score 1 · Answer 2 · answered Jun 26 '18 at 16:37

1

No, take $\mathbb{Z}^{\times \mathbb{N}}$ and the subgroup $(n,0,...0)$

answered Jun 26 '18 at 16:37

Tsemo Aristide

87,475

If a group and its quotient group are isomorphic, the normal subgroup only has identity element?

2 Answers2