# Homomorphism from $S_3$ to ($\mathbb{Q},+)$

I am solving exercise in abstract algebra and could not solve this 1 correctly.

Does there exists a homomorphism from $$S_3$$ to the additive group ($$\mathbb{Q},+)$$ of rational numbers?

I think it exists. Map $$A_3$$ to $$1$$ and remaining elements to $$-1$$. But answer is no.

So, what mistake I am making? Please tell.

• Your map is a homomorphism to the multiplicative group of non-zero rational numbers Nov 6 '20 at 17:36
• The answer should be "yes" since there is always the trivial homomorphism $x \mapsto 0$ Nov 6 '20 at 17:36
• And in fact, the trivial homomorphism is the only homomorphism that exists. Nov 6 '20 at 17:43

$$\{0\}$$ is the only finite subgroup of $$(\Bbb Q,+)$$, and hence there is no nontrivial homomorphism from any finite group $$G$$ to $$(\Bbb Q,+)$$.

• Does Image of $S_3$ need to be finite ? Why ? Nov 10 '20 at 16:27
• @Avenger, for any map $f\colon A \to B$, $a\mapsto f(a)$, the map $\tilde f\colon A\to f(A)$, $a\mapsto \tilde f(a):=f(a)$, is surjective, and hence $|f(A)|\le |A|$. Therefore, if $A$ is finite, $f(A)$ is finite as well.
– user810157
Nov 10 '20 at 16:58

Your map is a homomorphism to the multiplicative group of non-zero rational numbers.

You could map every element of $$S_3$$ to $$0$$ in $$\mathbb Q$$

to obtain a homomorphism to the additive group of rational numbers.

• Is it the only one?
– user810157
Nov 6 '20 at 20:13
• Yes, as commented above by Daniel Schepler Nov 6 '20 at 20:19