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

Sketch of my answer:

$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.

Please comment/correct my answer.

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    A sylow 2-subgroup has order 8.2012-10-05
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    So, two distinct Sylow 2-subgroups of $S_5$ could be $\langle (1234)\rangle$ and $\langle (1432) \rangle$.2012-10-05
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    $\langle(1234)\rangle=\langle(1432)\rangle$ 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 $16\not\mid5!$.2012-10-05
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    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 $S_5$.2012-10-05
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    Yes I see that anon. Thanks. I'm still confused as to why the order is 8. I thought we need $n_2= 1 \mod 2$ and $n_2 \| m$. Doesn't $n_2$ represent the number of Sylow p-subgroups?2012-10-05
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    It would help if you were to explain what the symbols $p, \alpha, m \text{ and } n_2$ represent - they are guessable. As a hint, how large are the Sylow subgroups of $S_4$? And note that every subgroup of $S_5$ of order 2 or 4 is contained in a Sylow 2-subgroup - so every Sylow 2-subgroup will contain a representative of each conjugacy class of elements of order 2 or 4.2012-10-05
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    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.)2012-10-05
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    anon- Thanks, that was silly of me!2012-10-05

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