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Let $G$ be a finite group such that Sylow $p$-subgroup $G$ has order $p$. Let $x$ and $y$ be two elements of order $p$. Is true that $x$ and $y$ are conjugate in $G$?

User1257
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This is not true. The smallest counterexample is a cyclic group of three elements. For example in the group $\Bbb{Z}/3\Bbb{Z}$ the cosets $\overline{1}$ and $\overline{2}$ are both of order $3$, but they are not conjugate.

Jyrki Lahtonen
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    It's also rarely true in non-trivial cases. Since the Sylow $p$-subgroup $P$ is abelian, it suffices to consider conjugacy in its normalizer (so we can assume the Sylow $p$-subgroup is normal). The conjugacy action depends only on $N_G(P)/C_G(P)$ which in this case is a cyclic group of order dividing $p-1$. The condition that $x$ and $y$ of order $p$ are always conjugate is equivalent to $[N_G(P):C_G(P)]=p-1$. – Jack Schmidt Jun 24 '14 at 15:09
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    Quite. An example where the situation described by @JackSchmidt occurs is $p=5$, $G=A_5$. The group $A_5$ has 24 elements of order $5$, hence six Sylow $5$-subgroups. If $P$ is one of them, then $|N_G(P)|=10$. 24 $5$-cycles $\implies|C_G(P)|=5$ and $[N_G(P):C_G(P)]=2$. Consequently the 5-cycles of $A_5$ are partitioned into $(p-1)/2=2$ conjugacy classes. A fact many here have encountered in a course on representation theory. – Jyrki Lahtonen Jun 24 '14 at 18:43