A group of order $150$ has at least $4$ conjugacy classes made by elements of order a power of $5$? Let be $G$ a group of order $150$ and $g \in G\,\,$ s.t. $|g|=25\,$:
Prove that $\exists \,\,K\lhd G\,\,$s.t. $K \simeq C_5$. $\,\,$ Say whether $\exists \,\,h \in G\,\,$ s.t. $|h| =15$.$\,\,$ Prove that if $\,\,\exists\,\, t\in G/K$ s.t. $\,\,|t|=15$ then $G/K$ has only one $5$-Sylow; in this case how many elements of order $15$ does $G$ have? $\,$Prove that $G$ has at least $4$ conjugacy classes made by elements of order a power of $5$.
I am struggling with the last part of this exercise, any help will be greatly appreciated. My solution:
We know, by hypothesis, that $\exists \,\,g\in G$ s.t. $|g|=25\,\,$ and this means that $C_{25} \leqslant G\,$, where $\langle g \rangle = C_{25}\,$, which is abelian because $25$ is the square of a prime, but it is also a $5$-Sylow of $G$, because $150=2 \cdot 3 \cdot 5^2$, hence, for the Sylow's Theorems, the conjugacy action of $G$ on $C_{25}$ is transitive, which implies $C_{25} \lhd G\,\,$. Observing that $g^5 \in C_{25}$ and that $|g^5|=5$ in $C_{25}$, we have $C_5 \simeq \langle g^5 \rangle \lhd C_{25}\,\, \Rightarrow \,\, C_5 \lhd G$.
$G$ has an element $h$ of order $15 \iff C_{15}\leqslant G$. Let be $P \in{\rm Sylow}_3(G)$, now $|P| = 3$ which is a prime, therefore $P \simeq C_3$. We have seen that $C_5 \lhd G$, thus $H := C_3C_5 \leqslant G$. From this: $C_3 \cap C_5 = \{1\}$, $|C_3C_5| = 15 = |H|$ and $C_3, C_5 \lhd H$ because they are the only $3$-Sylow and the only $5$-Sylow in $H$; hence $H \simeq C_3 \times C_5 \simeq C_{15} \leqslant G$. $h$ is the generator of $C_{15}$.
By hypothesis $C_{15} \leqslant G/K$, with $K \simeq C_{25}$. Now $|G/K|=30 =3 \cdot 2 \cdot 5$ and $[G/K : C_{15}]=2$, which is the minimum prime that divides the order of $G/K$, that implies $C_{15} \lhd G/K\,\,$. We can observe that $C_{15}$ has only one $5$-Sylow $Q$, hence $Q \lhd C_{15} \lhd G/K$, therefore $Q \lhd G/K$, which means that $Q$ is the only $5$-Sylow of $G/K$.
Any idea about the last two requests? Thank you.
 A: This is just a hint for the final question. Since the order of the group is not divisible by $4$ (the order of ${\rm Aut}(C_5)$), there must be at least two conjugacy classes of elements of order $5$ that lie in $K$.
Similarly, if $g$ has order $25$, then by considering $\langle g \rangle/K$, there must be at least two conjugacy classes of elements of order $25$ that lie in $\langle g \rangle$.
A: I'm trying to give a “canonical” answer requested by AleVanDerBauch. At the same time I will correct some mistakes in F.inc's arguments.
First of all, it's not so easy to conclude that there is only one Sylow 5-subgroup (by Sylow's theorem, also six Sylows are conceivable). Anyway, let $P=\langle x\rangle$ be a (cyclic) Sylow $5$-subgroup of $G$. Then $G$ acts on the set of (left) cosets of $P$ by (left) multiplication. This gives a homomorphism $f\colon G\to S_6$ (symmetric group) whose kernel is contained in $P$ since it fixes the trivial coset. Since $S_6$ has no element of order $25$, $f$ cannot be injective. Hence $K=\langle x^5\rangle$ is contained in  $\ker f$. Since $K$ is the unique minimal subgroup of $P$, we must have $K\unlhd G$ (in the OP it is wrongly written that $K\cong C_{15}$).
Now consider $G/K$ (a group of order 30). If $P$ is not normal, then $G/K$ has six Sylow $5$-subgroups which contain $6\cdot 4+1=25$ elements in total. This leaves only five elements of order $2$ or $3$. Hence, $G/K$ has a normal subgroup $H/K$ of order $6$. However, $P/K$ can only act trivially on $H/K$ (it cannot permute elements of order 2 or 3 non-trivially). Hence, $P/K$ centralizes $H/K$ and this leads to the contradiction $P\unlhd G$. We have therefore shown that $P\unlhd G$. Now any element $tK\in G/K$ of order $3$ acts trivially on $P/K$. Consequently, $txK$ is an element of order $15$ (note that we do not need to assume the existence of such an element).
Observe that $P\langle t\rangle\unlhd G$ as a subgroup of index $2$. Every element of order $15$ lies in $\langle x^5t\rangle$. The number of those is $\varphi(15)=8$.
Finally, I pick up Derek's answer to the last question. Suppose that $g\in G$ satisfies $gxg^{-1}=x^2$. Then $g^2xg^{-2}=x^{-1}$. Hence, $4$ divides the order of $g$. This contradicts Lagrange's theorem. Therefore, $x$ and $x^2$ are not conjugate in $G$. For the same reason, $x^5$ and $X^{10}$ are not conjugate. Clearly, $x$ and $x^5$ are not conjugate. So we have found four non-conjugate $5$-elements (the identitiy element is another $5$-element by the way).
