$G$ is of order 1225. According to the Sylow theorem, we know that $H$ is the only sylow 5-subgroup of $G$ and $K$ is the only sylow 7-subgroup of $G$. I am trying to prove that they are both normal but looking at the subgroup $gHg^{-1}$ which has the same order of H. What's next shall I look at? And how can we prove that $H$$\cap$$K$ = {e}? Could anyone give me a hint? Thanks!
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Part of the Sylow theorem also says that all p-Sylow subgroups are conjugate. From this you can get that a unique Sylow p-subgroup is normal, as well as characteristic. And, these 3 conditions are equivalent for p-Sylows. For the intersection, as Dylan's comment says, the intersection is a subgroup and must divide the orders of $H$ and $K$. |
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