The full question is as follows:
Let $G$ be a group of order $3\cdot5^2\cdot7$.
(a) Let $P$ be a subgroup of $G$ of order $7$. Show that $P$ is normal, or that $G$ contains a cyclic subgroup $C$ of order $35$.
(b) Assuming such a $C$ exists, let $x$ be an element of $C$ of order $5$. Show that the normalizer of $\langle x\rangle$ in $G$ has order equal to a multiple of $5^2\cdot7$.
(c) Explain why the above shows that $G$ is not simple.
Here's what I have:
(a) By the Sylow counting theorem, the number of Sylow 7-subgroups (denoted $n_7$) is congruent to 1 (mod 7). The possibilities are $n_7=1$ or $n_7=15$. If $n_7=1$, then the Sylow 7-subgroup is normal. Otherwise, $n_7=15$ and it follows from the $n!$ theorem that $|n_7|=|G:N_G(S)|$, for some $S\in Syl_7(G)$. By LaGrange's theorem, this gives that the normalizer of $S$ in $G$ is a subgroup of order 35. It remains to be shown that this group is cyclic. This is straightforward application of a theorem that states if $|G|=pq$ for primes $p<q$, and $p\not|(q-1)$, then $G$ is cyclic.
(b) I'm not sure what to do for this. I'd like to use the "normalizers grow theorem", that is, given a finite $p$-group $G$ and proper subgroup $H$, then $H\subsetneq N_G(H)$. However, in this case, $G$ is not a $p$-group. Any advice on how to approach this would be greatly appreciated.
(c) By part (b) and LaGrange's theorem, $|N_G(\langle x\rangle)|=5^2\cdot7$ or $3\cdot5^2\cdot7$. If the latter is true, then $\langle x\rangle$ is normal in $G$ (see https://proofwiki.org/wiki/Normal_Subgroup_iff_Normalizer_is_Group), so $G$ is not simple. If the former is true, then $|[G:N_G(\langle x\rangle)]|=3$. Because $3$ is the smallest prime divisor of $G$, $N_G(\langle x\rangle)$ is normal in $G$, and hence $G$ is not simple.
For part (b) you know that $\langle x\rangle$ is normal in $C$, but you also know that there is a $5$-Sylow subgroup Q in which $\langle x\rangle$ is normal. Let's see that $\langle x \rangle$ is normal in the group generated by $C$ and $Q$. If you take some $c\in C$ and some $t\in Q$, then $ct\langle x\rangle t^{1}c^{-1}=\langle x\rangle$, you can generalize this argument for any element of $\langle C, Q\rangle$.
Since the normalizer of $\langle x\rangle $ is the largest subgroup in which $\langle x\rangle$ is normal, it has to contain $\langle C, Q\rangle$, and this implies, by Lagrange theorem, that $35$ and $5^2$ are divisors of $N_{G}(\langle x\rangle)$.
I hope my answer can help you.