Proving $\{gY:g\in X\}=\{Yg:g\in X\}$

Question

I was trying the study a new subject in groups (from a book I have). I came acorss with the following question:

Consider:

$X=S_4$

$Y=\{e,(12)(34),(13)(24),(14)(23)\}$

How can I prove that:

$$\{gY:g\in X\}=\{Yg:g\in X\}$$

Answer 1

To prove $GY \subset YG$, you need to prove that for all $g_1 \in G$ and $y_1 \in Y$, there exists $y_2 \in G$ and $y_2 \in Y$ such that $g_1 y_1 = y_2 g_2$, which is equivalent to saying such that $g_1 y_1 g_2^{-1} \in Y$. Proving the other inclusion is similar.

We can actually prove a stronger property : for all $g \in G$ and $y \in Y$, we have $gyg^{-1} \in Y$.

Indeed, if $y=e$ we have $gyg^{-1}=e\in Y$. Otherwise, $y$ is of the form $y=(a \, b) (c \, d)$ (with $a, b, c, d$ numbers from $1$ to $4$ (in some order). We compute that

$$g y g^{-1} = (g(a) \, g(b)) (g(c) \, g(d))$$

And we check that this is still in $Y$ (note that $Y$ contains all permutations of the form $(a \, b) (c \, d)$, because for example $(12)(34)=(21)(34)=(21)(43)=(34)(12)=(43)(12)=\ldots$).