## Compute the number of Sylow subgroups in Sym(5) and Alt(5)

For $p \in \{ 2,3,5 \}$ find $n_p(A_5)$ and $n_p(S_5)$.

We directly computed the subgroup lattice of $A_4$, so we know that $n_2(A_4) = 1$ and $n_3(A_4) = 4$.

Note that $|A_5| = 60 = 2^2 \cdot 3 \cdot 5$. By Sylow’s theorem, we have the following.

• $n_2(A_5) \in \{ 1,3,5,15 \}$
• $n_3(A_5) \in \{ 1,4,10 \}$
• $n_5(A_5) \in \{ 1,6 \}$

We know that $A_5$ is simple, so $n_5(A_5) = 6$. Now the Sylow 3-subgroups of $A_5$ have order 3; by this previous exercise, $S_5$ has 20 elements of order 3, all of which are in $A_5$. Each of these is in a Sylow 3-subgroup of $A_5$, and each Sylow 3-subgroup is generated by two elements. Thus $n_3(A_5) = 10$. Note that the elements of order 4 in $S_5$ are precisely the 4-cycles, and none of these are contained in $A_5$. Thus the nonidentity elements of any Sylow 2-subgroup of $A_5$ have order 2; these elements are precisely the products of two 2-cycles. By this previous exercise, $S_5$ contains 15 products of two disjoint 2-cycles, all of which are contained in $A_5$ and each of which is contained in some Sylow 2-subgroup. Note that the Sylow 2-subgroups of $A_5$ are abelian, and that every element of order 2 in $A_5$ fixes exactly one element in $\{1,2,3,4,5\}$. Let $\sigma,\tau \in A_5$ such that $\sigma$ fixes $a$ and $\tau$ does not fix $a$. If $\sigma\tau = \tau\sigma$, then $\tau(a) = \sigma(\tau(a))$; becuase $\sigma$ fixes only $a$, we have $\tau(a) = a$, a contradiction. So $\sigma \tau \neq \tau \sigma$. Thus, if $\sigma,\tau \in A_5$ are elements of order 2 that do not fix the same element, then no Sylow 2-subgroup contains both $\sigma$ and $\tau$. For each $k \in \{1,2,3,4,5\}$, there are precisely three products of two 2-cycles in $A_5$ which fix $k$, corresponding to the image of $V_4$ (in $A_4$) under a renaming injection $A_4 \rightarrow A_5$, and each such image is a Sylow 2-subgroup. Thus $n_2(A_5) = 5$.

Now consider $S_5$; note that $|S_5| = 120 = 2^3 \cdot 3 \cdot 5$. By Sylow’s Theorem, we have the following.

• $n_2(S_5) \in \{ 1,3,5,15 \}$
• $n_3(S_5) \in \{ 1,4,10,40 \}$
• $n_5(S_5) \in \{ 1,6 \}$

Every Sylow 5-subgroup of $A_5$ is a Sylow 5-subgroup of $S_5$, so $n_5(S_5) = 6$. Similarly, every Sylow 3-subgroup of $A_5$ is a Sylow 3-subgroup of $S_5$, so that $n_3(S_5) \geq 10$. Now the nonidentity elements in Sylow 3-subgroups of $S_5$ are precisely the 3-cycles, and all of these are contained in $A_5$. So $S_5$ has no additional Sylow 3-subgroups; hence $n_3(S_5) = 10$. Note that $S_4 \leq S_5$, and that every Sylow 2-subgroup of $S_4$ is a Sylow 2-subgroup of $S_5$. In this previous exercise, we saw that $H_1 = \langle (1\ 2\ 3\ 4), (2\ 4) \rangle$, $H_2 = \langle (1\ 3\ 2\ 4), (3\ 4) \rangle$, and $H_3 = \langle (1\ 2\ 4\ 3), (2\ 3) \rangle$ are the Sylow 2-subgroups of $S_4$. Then $H_4 = \langle (1\ 2\ 3\ 5), (2\ 5) \rangle$, $H_5 = \langle (1\ 3\ 2\ 5), (3\ 5) \rangle$, and $H_6 = \langle (1\ 2\ 5\ 3), (2\ 3) \rangle$ are also distinct Sylow 2-subgroups of $S_5$, and clearly $H_1$, $H_2$, and $H_3$ are pairwise distinct from $H_4$, $H_5$, and $H_6$. (In particular, the former fix 5 and the latter do not.) So $n_2(S_5) \geq 6$; thus $n_2(S_5) = 15$.

Typo: Thus, if $\sigma, \tau \in A_4$ are elements of order 2 – $A_4$ should be $A_5$.
Also we can directly notice $n_2(S_5)=15$, since there are 5 ways of embedding $S_4$ into $S_5$, so $n_2(S_4) \cdot 5 = 15$.
Thanks! Also, that argument for $n_2(S_5)$ is slick.