How can I find left class residue of $H=\{\sigma \in S_4\mid \sigma (4)=4\}$ ? Same question for $H'=\{\sigma \in S_n\mid \sigma (n)=n\}$.
Attempt
First of all, all element of $S_4$ that fix $4$ are in the class of $1$. But then, I tried by hand to compute all $\sigma ^{-1}\tau$ for $\tau $ and $\sigma $ in $H$ and see if they are in $H$. The problem it's very long, so is there an other technic ?
Suppose $\sigma(4) = \tau(4) = a$, where $a \in \{1,2,3, 4\}$.
Then $\sigma^{-1}(a) = 4$, and thus $\sigma^{-1}\tau(4) = \sigma^{-1}(\tau(4)) = \sigma^{-1}(a) = 4$, hence $\sigma^{-1}\tau \in H$, and $\sigma H = \tau H$.
On the other hand, if $\sigma H = \tau H$, then $\sigma = \tau\rho$, for some $\rho \in H$, whence $\sigma(4) = \tau\rho(4) = \tau(\rho(4)) = \tau(4)$.
It follows that a left transversal of $H$ in $S_4$ is:
$\{e,(3\ 4), (2\ 4), (1\ 4)\}$, that is, the left cosets are:
$H = \{e, (1\ 2\ 3), (1\ 3\ 2), (1\ 2), (1\ 3), (2\ 3)\}$
$(3\ 4)H = \{(3\ 4), (1\ 2\ 4\ 3), (1\ 4\ 3\ 2), (1\ 2)(3\ 4), (1\ 4\ 3), (2\ 4\ 3)\}$
$(2\ 4)H = \{(2\ 4), (1\ 4\ 2\ 3), (1\ 3\ 4\ 2), (1\ 4\ 2), (1\ 3)(2\ 4), (2\ 3\ 4)\}$
$(1\ 4)H = \{(1\ 4), (1\ 2\ 3\ 4), (1\ 3\ 2\ 4), (1\ 2\ 4), (1\ 3\ 4), (1\ 4)(2\ 3)\}$
which only takes a few minutes to compute.