Quotient group $G/H$ isomorphic to $G$

Question

I was wondering if the following statement is true or not. Let $G$ a group and $H \triangleleft G$

$$ G/H \cong G \implies H = \left \{ 1 \right \}$$

I know it is true if $G$ is a finitely generated abelian group, as you can see here. Nevertheless, my intuition says it should be not true in the general case. Actually, I was trying to find a counterexample, taking G as a free group. However, I haven't found anything.

This is a duplicate of this old question. (Incidentally, taking $G$ as a finitely generated free group cannot work. One reason for this is because free groups are residually finite, and finitely generated, residually finite groups cannot have this property, by a theorem of Mal'cev.) — user1729, Feb 22 '15 at 18:24

score 2 · Accepted Answer · answered Feb 22 '15 at 18:21

2

You could let $G$ be the free group on countably many letters, and let $H$ be the normal subgroup generated by the first letter. Then $G/H$ is isomorphic to the free group on countably many letters, which is isomorphic to $G$.

answered Feb 22 '15 at 18:21

hunter

29,847

(of course, the natural map $G \to G/H$ will not be an isomorphism unless $H$ is trivial) – hunter Feb 22 '15 at 18:22

score 2 · Answer 2 · answered Feb 22 '15 at 18:22

2

A counterexample is given by the multiplicative group of the unit circle in the complex plane. The quotient by the subgroup $\{-1,1\}$ is isomorphic to the circle.

answered Feb 22 '15 at 18:22

Matt Samuel

58,164

Quotient group $G/H$ isomorphic to $G$

2 Answers2

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