Conjugate subgroups and conjugate elements

Question

While trying to prove that the alternating group $A_5$ is a simple group, I came across two assertions I see as contradicting, that is :

• the 5-cycles are not all conjugate to each other

(proven here : Show that not all 5 cycles in $A_5$ are conjugate in $A_5$)

• if $\sigma$ and $\sigma'$ are 5-cyles, then by one of the Sylow theorems, $<\sigma>$, which is a 5-Sylow is conjuguate to $<\sigma'>$, another 5-Sylow

Can anyone demystify this ?

Answer 1

Two cyclic groups are conjugate if each generator of one is conjugate to a generator of the other.

This does not mean that every generator of one is conjugate to every generator of the other.

In particular if $A=\langle a\rangle$ and $B=\langle b\rangle$ then $a\sim b^n$ and $b\sim a^m$ for some $n,m$ relatively prime to the orders of $a$ and $b$ (so that $b^n$ and $a^m$ are also generators of $B$ and $A$ respectively), but from this we cannot deduce that $a\sim b$, or in other words we still don't know if $a,b$ are themselves conjugate.

Answer 2

More specific to the alternating groups, when single conjugacy classes in $S_n$ split into two classes in $A_n$, we have $a\not\sim a^{-1}$, however $\langle a\rangle=\langle a^{-1}\rangle$ is always true.