# Does there exist non trivial group homomorphism from $S_3$ to ( $\mathbb Q$,+)

Let $G = S_3$ be the permutatiin group of 3 symbols.Then

1. $G$ is isomorphic to a subgroup of a cyclic group
2. There exists a cyclic group $H$ such that $G$ maps homomorphically onto $H$.
3. $G$ is a product of cyclic groups
4. there exists a nontrivial group homomorphism from $G$ to the additive group ($\mathbb Q$,+) of rational numbers

1 option is clearly false since subgroup of a cyclic group is again cyclic and $G$ can't be isomorphic to a cyclic group

2 option is true since there is an epimorphism from $G$ onto $\mathbb Z_2$

3 option is false since $G$ is non commutative

Now I only get trivial homomorphism from $G$ to additive group of rational numbers. So option 4 is false. Am I right?

• Re 2, note that the trivial group is also cyclic :) -- In $\Bbb Q$, all elements $\ne0$ have infinite order. Hence for any finte group, there is only the trivial homomorphism to $\Bbb Q$. Jun 19 '18 at 9:51
• But Option 3 is true. $G$ is the product $HK$ with $|H|=3$ and $|K|=2$. Jun 19 '18 at 10:46

The option 4. is false because $(\mathbb{Q},+)$ has no elements of finite order other than $0$, whereas every element of $S_3$ has finite order.
• @JoséCarlosSantos ...except the identity element $0$. Jun 20 '18 at 9:19