Without having discussed cyclic groups yet, in my textbook I find statements about isomorphisms like this:

$$ 2\mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} $$


$$ S_n/A_n \cong \mathbb{Z}/2\mathbb{Z} $$

with $S_n$ being the symmetric group and $A_n$ the alternating group. I don't understand how to get to these statements. Can anyone help?

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    $\begingroup$ Do you know the first isomorphism theorem ? $\endgroup$ – Arnaud D. Jan 9 '20 at 16:58
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    $\begingroup$ You haven't learned about cyclic groups but you have studied symmetric and alternating groups? That seems like an unusual order of topics. What do you know about $S_n, A_n$? $\endgroup$ – lulu Jan 9 '20 at 17:02
  • $\begingroup$ @ArnaudD. yes, I know it but the book hadn't mentioned it yet. My guess is that one should be able to show that there is an isomorphism and definie that isomorphism concret. $\endgroup$ – KingDingeling Jan 9 '20 at 17:13
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    $\begingroup$ In each of the two equations you give, on the left, $G/H$ has only two elements, namely, $H$ and the unique nontrivial coset of $H$ in $G$, while on the right, $2Z$ has just two elements, namely, $2Z$ itself, and the coset containing $1$, that is to say, the coset consisting of the odd integers. Now, it's pretty easy to set up an isomorphism between two groups of order two, as the identity must map to the identity, so the nonidentity must map to the nonidentity. That tells you how to construct the isomorphisms, now you just have to show that they actually are isomorphisms. $\endgroup$ – Gerry Myerson Jan 9 '20 at 18:04
  • $\begingroup$ Which textbook are you referring to? $\endgroup$ – Shaun Jan 9 '20 at 18:59

One can summarise the comments that you'll need some basic material on group homomorphisms and quotient groups. For a group homomorphism $\phi\colon G\rightarrow H$ we have $$ G/\ker(\phi)\cong \phi(G)\subseteq H. $$ If we take the signum homomorphism from $S_n$ to $C_2$ we obtain $\ker(\phi)=A_n$ by definition so that $$ S_n/A_n\cong C_2. $$ Here $C_2$ is the unique group of order $2$. It is cyclic and additively written it is $\Bbb Z/2\Bbb Z$.


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