There is no non-trivial homomorphism $\mathbb{Q}\rightarrow S_3$

Question

I want to show that the following problem is true :

Please help me to show it.

Thanks.

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Hint: $\mathbb{Q}$ is divisible so there is no nontrivial subgroup of finite index (see here for example).

edited Apr 13 '17 at 12:20

Community

1

answered Feb 17 '13 at 09:51

Seirios

33,157

Thanks for remarking that points. +1 – Mikasa Mar 23 '13 at 19:30

Mikasa · Answer 2 · 2013-02-17T11:43:57.740

4

$\mathbb Q$ is divisible so every quotient of it is divisible also. But if $G$ is abelian finitely generated or finite, it can't be divisible. This can be another but a bit similar to @Seirios approach.

edited Feb 17 '13 at 11:43

answered Feb 17 '13 at 10:00

Mikasa

67,374

If $G$ is finitely generated and abelian, it can't be divisible; I think there are examples of non abelian finitely generated divisible groups. – Seirios Feb 17 '13 at 10:08
@Seirios: Yes you are right. I meant abelian ones and we are working on Q here. I 'll add this point. Thanks – Mikasa Feb 17 '13 at 11:43
Nice point :-) +1 – amWhy Feb 18 '13 at 00:06

There is no non-trivial homomorphism $\mathbb{Q}\rightarrow S_3$

2 Answers2