Let $X = \{1, \dots, n\}$. Let $G$ act on $X$, denoted by $g \cdot x$ for $g \in G$ and $x \in X$. Let $K$ be the kernel of this action.

  1. Show that the map $(gK) * x = g \cdot x$ gives a well-defined action of $G/K$ on $X$.
  2. Show that $\varphi: G/K \to S_n$ given by $[\varphi(gK)](x) = (gK) * x$ defines an injective homomorphism.
  3. Let $G$ be the rotation group of the cube. We know that $G \cong S_4$, by letting $S_4$ act on the set of pairs of opposite faces of the cube. Consider the action of $G$ on the set of pairs of opposite faces on the cube. Find the kernel of this action, described as a subgroup of $S_4$. Find $S_4/K$.

I understand for $(1)$, we show that the map is an action and that it's well-defined. I also understand for $(2)$, we show that the map is a homomorphism and that it's injective. But I'm stuck otherwise, and any help would be appreciated.



If $a, b$ are in the same left coset of $K$, $a^{-1}b = k$ for some $k \in K$. Hence $b \cdot x = ak \cdot x = a \cdot x$ since any element of the kernel fixes every $x \in X$. Thus this is a well-defined action.


$$\varphi(ghK)(x) = gh \cdot x = g \cdot (h \cdot x) = \varphi(gK)(h \cdot x) = \varphi(gK)(\varphi(hK)(x))$$so $\varphi$ is a homomorphism. If $\varphi(gK)(x) = \varphi(hK)(x)$ for all $x \in X$, then $g \cdot x = h \cdot x$, which means $g^{-1}h \in K$, which in turn means $gK= hK$. Hence $\varphi$ is injective.


$K$ is the Klein four-group $($opposite faces remain fixed if and only if $180^\circ$ rotation about center, hence group of order $4$ with all elements of order $2$$)$. By $(2)$ we have an injective homomorphism $S_4/K \to S_3$. Since $|S_4/K| = 6$, $S_4/K \cong S_3$.


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