Are the Sylow $p$-subgroups of $S_4$ also Sylow $p$-subgroups of $S_5$?

Question

Here is my thinking:

$|S_4| = 4! = 1 \times 2 \times 3 \times 4 = 2^3 \times 3$.

$|S_5| = 5! = 1 \times 2 \times 3 \times 4 \times 5 = 2^3 \times 3 \times 5$.

Since $2^3$ is the maximal power of $2$ which divides the order of $S_4$ and $S_5$ and since $S_4$ is a subgroup of $S_5$, we know that the Sylow $2$-subgroups of $S_4$ are also Sylow $2$-subgroups of $S_5$.

Similarly, since $3$ is the maximal power of $3$ which divides the order of $S_4$ and $S_5$ and since $S_4$ is a subgroup of $S_5$, we know that the Sylow $3$-subgroups of $S_4$ are also Sylow $3$-subgroups of $S_5$.

On the other hand, if we wanted to check whether the Sylow $p$-subgroups of $S_4$ are also Sylow $p$-subgroups of $S_6$ we would see the following.

$|S_4| = 4! = 1 \times 2 \times 3 \times 4 = 2^3 \times 3$.

$|S_6| = 6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 2^3 \times 3 \times 6 = 2^4 \times 3^2$.

Since $6$ can be decomposed into $2 \times 3$, the maximal powers of $2$ and $3$ are no longer the same in $S_4$ and $S_6$ and so the Sylow $p$-subgroups of $S_4$ are not also $p$-subgroups of $S_6$.

Is this correct?

Answer 1

Hint: Embed $S_4\hookrightarrow S_5$ and then compose with the inclusion $P\hookrightarrow S_4$. Once you've a subgroup of the right order, it's a Sylow subgroup!