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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?

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    $\begingroup$ 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 $\Bbb Q$. $\endgroup$ – Hagen von Eitzen Jun 19 '18 at 9:51
  • $\begingroup$ But Option 3 is true. $G$ is the product $HK$ with $|H|=3$ and $|K|=2$. $\endgroup$ – Derek Holt Jun 19 '18 at 10:46
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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.

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  • $\begingroup$ Thank you sir .Are my other options right? $\endgroup$ – Normal Jun 19 '18 at 9:58
  • $\begingroup$ @Normal They are right and your explanations are just fine. $\endgroup$ – José Carlos Santos Jun 19 '18 at 9:59
  • $\begingroup$ @JoséCarlosSantos ...except the identity element $0$. $\endgroup$ – Leppala Jun 20 '18 at 9:19
  • $\begingroup$ @Leppala I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jun 20 '18 at 9:35

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