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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 ?

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  • $\begingroup$ The classes are determined by $\sigma(4)$, or by $\sigma(n)$ is the general case. $\endgroup$ – Derek Holt Jul 19 '17 at 9:07
  • $\begingroup$ Indeed, the classes are $\sigma_iH$ with $\sigma_i(n)=i$, $1\leq i\leq n$. $\endgroup$ – Wuestenfux Jul 19 '17 at 9:16
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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.

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