I need to find the Sylow $p$-subgroups of the alternating group $A_5.$ So I need to find the maximal $p$-subgroups of $A_5.$
First of all, what are the elements of $A_5$? I know they are the even permutations of $\{1,2,3,4,5\}$ but what does this mean?
Apparently $\mathbb Z/3\mathbb Z= \left<(1,2,3)\right> $ is a Sylow $3$- subgroup of $A_5.$ Why is this?
There are several ways of defining the sign $sgn:S_n\to \{-1,1\}$ of a permutation. The key property of $sgn$ is that $sgn(\sigma)=-1$ if $\sigma$ is a transposition, that is, a permutation that exchanges two elements. $A_n$ is defined to be the kernel of $sgn$,i.e. is set of permutations which have sign 1 (also called even permutations). A sylow p-subgroup of a finite group $G$ is a subgroup $H$ such that $|H|=$ the largest power of p dividing $|G|$. Since $S_5=120$ we have $A_5=60=2^2\cdot 3\cdot 5$. $Z/3Z$ isn't exactly $<(1,2,3)>$ but is isomorphic to that subgroup since both are cyclic of order $3$, so $Z/3Z$ is isomorphic to a sylow 3-subgroup of $A_5$.