# Exhibit two distinct Sylow 2-subgroups of $S_5$.

Exhibit two distinct Sylow 2-subgroups of $S_5$ and an element of $S_5$ that conjugates one into the other.

$p=2,\, \alpha=3,\, m=15, n_2=1,3,5,15$ and so the number of Sylow 2-subgroups of $S_5$ are $15.$ Two distinct Sylow 2-subgroups of $S_5$ are $<(12)>$ and $<(23)>.$ Observe that $(123)\in S_5$ and $(123)(12)(132)=(23).$ So $(123)$ is the element that conjugates one into the other.

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A sylow 2-subgroup has order 8. –  i. m. soloveichik Oct 5 '12 at 15:24
So, two distinct Sylow 2-subgroups of $S_5$ could be $\langle (1234)\rangle$ and $\langle (1432) \rangle$. –  Lyapunov Oct 5 '12 at 15:33
$\langle(1234)\rangle=\langle(1432)\rangle$ because $(1432)=(1234)^{-1}$, so they are not distinct $2$-subgroups, and this cyclic subgroup is order $4$ instead of $8$ anyway. Note that what distinguishes $p$-Sylow subgroups from other $p$-subgroups is the former have maximal prime power order; which means order $8$ because $8\mid5!$ but $16\not\mid5!$. –  anon Oct 5 '12 at 15:35
Note that $S_5$ contains the point stabilizers, which are subgroups isomorphic to $S_4$. Since the Sylow $2$-subgroups of $S_5$ and $S_4$ have the same size, you are reduced to solve your problem for $S_4$ instead of $S_5$. –  Martino Oct 5 '12 at 15:40
You're confused as to why the order is 8, hmm... Do you know what the definition of p-Sylow subgroups is, firstly? Do you agree that 8 divides 5! but 16, the next highest power of 2, does not? Do you understand that the number of p-Sylow subgroups is not the same as the order (size/cardinality) of the individual p-Sylow subgroups? (The number of bank accounts I have is not equal to the dollar amounts in the accounts, after all.) –  anon Oct 5 '12 at 15:43

$K_1=<(1234), (24)>$
$K_2=<(1234), (34)>$.
$(23)K_1(23)=K_2$.