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$|A_5| = 60 = 2^2\times 3\times 5$.

The subgroup $H = \{1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$ is a Sylow 2-subgroup of $A_5$, $(1 2 3)$ is a Sylow 3-subgroups of $A_5$, and $(1 2 3 4 5)$ is a Sylow 5-subgroup of $A_5$. How should I find all $p$-Sylow subgroups and prove they are all the $p$-Sylow subgroups?

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2 Answers 2

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Technically, $(123)$ is not a $3$-Sylow subgroup, since it's not a subgroup; you mean the subgroup $(123)$ generates. The same for the $5$-Sylow subgroup.

To prove they are $p$-Sylow subgroups, verify that they are subgroups and that they have the correct size.

To find all $p$-Sylow subgroups for a given $p$, find one and use the fact that all $p$-Sylow subgroups are conjugate. For example, the set of all $3$-Sylow subgroups is the set of all conjugates of $\{1, (123), (132)\}$.

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By Sylow's theorems, the number of $2$-Sylow subgroups is $1$, $3$, $5$ or $15$. It cannot be $1$ or $3$ because $A_5$ is simple. If the number of $2$-Sylows is $15$, then the subgroup $H$ is self-normalizing, ie. $N_G(H) = H$. This isn't true, because $(123)$ normalizes $H$. Therefore there are five $2$-Sylows. Conjugate $H$ by different elements until you have found them all.

Finding the $3$-Sylow and $5$-Sylow subgroups is easier. They are cyclic, so you can find them by finding every element of order $3$ and order $5$ in $A_5$.

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