Let G be a group of order 105 and H be its subgroup of order 35. Then
A) H is a normal subgroup of G
B) H is cyclic
C) G is simple
D) H has a normal subgroup K of order 5 and K is normal in G
-35=5•7 and 5 doesn't divide (7-1) means H is cyclic.Hence it has a normal subgroup of order 5.(By sylow theorem)
But how to check if H is normal in G or not??
The group $G$ has order $|G|=105=3\cdot 5\cdot 7$. If $n_p$ is the number of Sylow-$p$ subgroups, then $n_3\in\{1,7\}$, $n_5\in\{1,21\}$, and $n_7\in\{1,15\}$.
By Cauchy's theorem, $G$ has subgroups $K_1$, $K_2$ of order $5$ and $7$ respectively. If neither of these are normal, then $n_5=21$ and $n_7=15$. But this gives us $175$ elements, which is impossible. Thus either $K_1$ or $K_2$ is normal, so $K_1K_2$ is a subgroup of order $\dfrac{|K||H|}{|H\cap K|}=35$.
Now write $H=K_1K_2$. Then $G$ acts transitively on the cosets $G/H$, so we have a map $G\to S_3$. If this map is trivial, then $H$ is normal in $G$. If not, then the image has order $3$ (since $|G|$ and $2$ are coprime), so the kernel is a normal subgroup of order $35$. If $x$ is in the kernel, then $xH=H$, so $x\in H$. That is, the kernel is exactly $H$, so again $H$ is normal in $G$.
From Sylow theory, $G$ acts transitively on all Sylow-$p$ subgroups via conjugation. But any conjugate of $K_1$ lies in $H$ since $H$ is normal. Since $H$ is necessarily cyclic, $K_1$ is normal in $H$, so $H$ contains $1$ Sylow-$5$ subgroup. This means $G$ also contains only $1$, and similarly for the Sylow-$7$ group.
Finally, note that it is not necessarily true that $n_3=1$, because there is a nonabelian group of order $105$. You can construct this from any nontrivial automorphism $C_3\to \operatorname{Aut}(C_5\times C_7)\cong (\Bbb Z/35)^\times$. One exists since $|(\Bbb Z/35)^\times|=(5-1)(7-1)=24=3\cdot 2^3$, so we can send a generator of $C_3$ to the element of order $3$.
