Let $|G|=5780$, prove $G$ has one and only subgroup of index $4$ Let $G$ be a group of order $5780$. prove that $G$ has one and only subgroup $H$ such that $[G:H]=4$, and by that conclude that $G$ has one and only 5-Sylow subgroup.
My Attempt:
$5780 = 2^2 \cdot 5 \cdot 17^2$. Using the first Sylow theorem we conclude that $G$ has a 5-sylow subgroup, and by the third Sylow theorem we conclude that the number of 5-sylow subgroups, let it be denotes $n_5$ is a divisor of $2^2 \cdot 17^2 = 1156$ and exists: $n_5  = 1 \mod 5$.
combining this arguments and we conclude: $n_5 \in \{1, 1156\}$. This doesn't really helped to determine that $n_5=1$, as I'd like, as if we assume $n_5=1156$ we won't have an immediate contradiction regarding the number of elements in $G$.
again, using the sylow theorem we conclude that we have 17-sylow subgroup and a subgroup of order $289$, but this doesn't really help us to determine that there exists a group $H$ of size $1145$, which means $[G:H]=4$.
 A: Consider $n_{17}$ the number of Sylow $17$ subgroups. So $n_{17}\mid 20$ and $\equiv 1\mod 17 $. So this forces $n_{17}=1$. Let $P_{17}$ be the unique normal Sylow $17$ sub-group. Then if $H$ is a subgroup of index $4$, then $P_{17}\subset H$. So you want a index $4$ subgroup of $G/P_{17}$ which is a group of order $20$ which is am as finding Sylow $5$ sub-group of $G/P_{17}$. But in a group of order $20$, the Sylow $5$ group is normal and hence there exists a unique subgroup of $G/P_{17}$ of index $4$. Then by the correspondence theorem, there exists a unique subgroup of index $4$ in $G$.
A: We know by Sylow's Theoroem that the Sylow $17-$subgroup is normal. We show each Sylow $q-$sungroup by $P_q$. So we know we have $P_ {17} $ and $P_5$. As $P_ {17}$ is normal in the group; thus, $P_ {17} P_5$ is a subgroup of the group. In this subgroup we have a Sylow $5-$subgroup, which is actually normal too. So $P_ {17} P_5$ is a subgroup of $N_G(P_5)$, and this means that $N_G(P_5)$ has at least $17^2.5$ elements. There are only two possibilities for $n_5$(number Sylow $5-$subgroups):
$$n_5= 1 \;or\; n_5=1156 \Rightarrow |N_G(P_5)|= |G| \;or\; 5 $$
If $n_5 \neq 1$, then the $N_G(P_5)$ has only five elements which is contradiction with the previous fact that $N_G(P_5)$ has at least $17^2.5$ elements. Therefor, $P_5$ is a normal subgroup in $G$.
