Finding cosets from symmetric subgroup $S_3$ of symmetric group $S_4$ where $S_3 = \{\phi\in S_4\mid\phi(4) = 4\}$

Question

This question confused me a bit.

Symmetric group $S_3$ is a subgroup of symmetric group $S_4$ where $S_3 = \{\phi \in S_4 \mid \phi(4) = 4\}$

We were told to find all the different cosets from $S_3$ using this theorem that said

$aH = bH$ if and only if $a^{-1} b \in H$

Our lecturer said that we need to find an element $\phi \in S_4$ but $\phi \notin S_3$ and later we got that the cosets are $S_3, \phi S_3, \phi^2 S_3$ and $\phi^3 S_3$.

I don't really understand, why it should be $\phi \in S_4$ but $\phi \notin S_3$? What would happen if $\phi \in S_3$?

Answer 1

By definition $\phi S_3$ means all products $\phi \sigma$ where $\sigma$ ranges over $S_3.$ So if $\phi$ is actually in $S_3$ this would just end up being only $S_3$ since multiplying all elements of a group by one of its elements just gives that group back. [a subgroup is in particular a group]

Answer 2

If $\phi\in S_3$, then $\phi S_3=S_3$.

There are four cosets. If we choose $\phi=(1234) $, say, then they are $S_3,\phi S_3,\phi^2S_3,\phi^3S_3$.

Not every $\phi\in S_4\setminus S_3$ works. For instance, $(24) $ has order two, so only gives two cosets.