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Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the $3$-adic expansion of $n.$ If $n = a.3^{m} +b$ where $a \in \{0,2 \},$ and $0 \leq b \leq 3^{m}-1,$ then the Sylow $3$-subgroup of $S_{n}$ is the direct product of $a$ copies of a Sylow $3$-subgroup of $S_{3^{m}}$ with a Sylow $3$-subgroup of $S_{b}.$ This reduces us to considering the case $n = 3^{m}.$ In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles interchanging the sets of points moved by the respective direct factors in a way compatible with their respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$

Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the base $3$ expansion of $n.$ If $n = a_{0} + 3 a_{1} + \ldots + 3^{m-1}a_{m-1}$ where each $a_{i} \in \{0,2 \}$, then a Sylow $3$-subgroup of $S_{n}$ is the direct product over $i$ of a direct product of $a_{i}$ copies the same as a Sylow $3$-subgroup of $S_{3^{i}}$. This reduces us to considering the case $n = 3^{m}$ for some $m$.

In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles interchanging the sets of points moved by the respective direct factors in a way compatible with their respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$

Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the $3$-adic expansion of $n.$ If $n = a.3^{m} +b$ where $a \in \{0,2 \},$ and $0 \leq b \leq 3^{m}-1,$ then the Sylow $3$-subgroup of $S_{n}$ is the direct product of $a$ copies of a Sylow $3$-subgroup of $S_{3^{m}}$ with a Sylow $3$-subgroup of $S_{b}.$ This reduces us to considering the case $n = 3^{m}.$ In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles permuting the points moved by the direct factors in a way compatible with the respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$

Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the $3$-adic expansion of $n.$ If $n = a.3^{m} +b$ where $a \in \{0,2 \},$ and $0 \leq b \leq 3^{m}-1,$ then the Sylow $3$-subgroup of $S_{n}$ is the direct product of $a$ copies of a Sylow $3$-subgroup of $S_{3^{m}}$ with a Sylow $3$-subgroup of $S_{b}.$ This reduces us to considering the case $n = 3^{m}.$ In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles interchanging the sets of points moved by the respective direct factors in a way compatible with their respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$

Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the $3$-adic expansion of $n.$ If $n = a.3^{m} +b$ where $a \in \{0,2 \},$ and $0 \leq b \leq 3^{m}-1,$ then the Sylow $3$-subgroup of $S_{n}$ is the direct product of $a$ copies of a Sylow $3$-subgroup of $S_{3^{m}}$ with a Sylow $3$-subgroup of $S_{b}.$ This reduces us to considering the case $n = 3^{m}.$ In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles permuting the points moved by the direct factors in a way compatible with the respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

Such a Sylow $3$-subgroup is a direct product of "iterated wreath products" of the cyclic group of order $3$. It depends on the $3$-adic expansion of $n.$ If $n = a.3^{m} +b$ where $a \in \{0,2 \},$ and $0 \leq b \leq 3^{m}-1,$ then the Sylow $3$-subgroup of $S_{n}$ is the direct product of $a$ copies of a Sylow $3$-subgroup of $S_{3^{m}}$ with a Sylow $3$-subgroup of $S_{b}.$ This reduces us to considering the case $n = 3^{m}.$ In that case, one takes a direct product of $3$ copies of a Sylow $3$-subgroup of $S_{3^{m-1}},$ and then adjoins a product of $3^{m-1}$ $3$-cycles permuting the points moved by the direct factors in a way compatible with the respective actions. This is more complicated to say than to implement: for example, to go from $S_{3}$ to $S_{9},$ take $\langle (123) \rangle \times \langle (456) \rangle \times \langle (789) \rangle$ and adjoin $(147)(258)(369).$

There is nothing in this construction which is particular to the prime $3$. The obvious analogues work for Sylow $p$-subgroups of $S_{n}$ for any prime $p.$