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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.

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    $\begingroup$ Your map is a homomorphism to the multiplicative group of non-zero rational numbers $\endgroup$ Nov 6 '20 at 17:36
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    $\begingroup$ The answer should be "yes" since there is always the trivial homomorphism $x \mapsto 0$ $\endgroup$
    – player3236
    Nov 6 '20 at 17:36
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    $\begingroup$ And in fact, the trivial homomorphism is the only homomorphism that exists. $\endgroup$ Nov 6 '20 at 17:43
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$\{0\}$ is the only finite subgroup of $(\Bbb Q,+)$, and hence there is no nontrivial homomorphism from any finite group $G$ to $(\Bbb Q,+)$.

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  • $\begingroup$ Does Image of $S_3$ need to be finite ? Why ? $\endgroup$
    – Avenger
    Nov 10 '20 at 16:27
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    $\begingroup$ @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. $\endgroup$
    – user810157
    Nov 10 '20 at 16:58
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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.

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

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