The isomorphism from $S_3/\langle (123)\rangle$ to $\mathbb{Z}_2$.

Question

Suppose that $N=\{(123), (132), \operatorname{e}\}$ and $N$ is normal in $S_3$. Show that the quotient group $S_3/N$ is isomorphic to $\mathbb{Z}_2$.

What mapping should I use?

Answer 1

First of all, if you already know that $N$ is normal, then you have that $S_3/N$ is a group of order $2$, and hence the only option is $\mathbb{Z}/2\mathbb{Z}$. If you want to give the explicit mapping that does that do the following:

From a comment I just saw, I think you dont know what the elements in $S_3/N$ are. Remember, $S_3/N$ is the quotient group, whose elements are the right cosets of $N$. Note that since $S_3$ has order $6$, and $N$ has order $3$, then you only need two cosets. Namely, $N$ and $xN$ for some $x$ not in $N$. Then the set $\{N,xN\}$ forms a group. Then you need to send $N\rightarrow 0$ and $xN\rightarrow 1$, to obtain your mapping.

Answer 2

Hint: The identity needs to be mapped to the identity. The other thing needs to be mapped to the other thing.

Answer 3

$S_3/N=\{N,(12)N\}$. Which one of these cosets is the identity under coset multiplication?

Answer 4

First, by using Lagrange's theorem, we know the index $[S_3:N]=|S|/3=6/3=2$, so that will be the order of the quotient group... Now the mapping between this and $\mathbb{Z}_2$ is trivial because, under isomorphisms, there exists only one group of order 2. I will let you realize that, just think about it, if you have a group of only two elements, not a lot of things can happen...