Let $G$ be a group of order $7105$, show that the order of the center of $G$ is divisible by $35$. Let $G$ be a group of order $5\cdot 7^2 \cdot 29$, show that the order of the center $Z(G)$ of $G$ is divisible by $35$.
Now, $G$ contains only a $5$-sylow, $P_5$, and since the conjugacy action of $G/P_5$ on $P_5$ is the trivial homomorphism, then $P_5 \subseteq Z(G)$, so $5$ divides the order of the center.
The number of $7$-sylow could be either $1$ or $29$, in the first case we can conclude as above. What about the second case?

This question is listed as unanswered. So I decided to post an answer. This $G$ has unique Sylow $29$-subgroup $S$ by the Sylow theorem. So $S$ is normal, $G/S$ has order $5\cdot 49=245$. $G/S$ has unique Sylow $5$-subgroup $T/S$ of order $5$ and unique $7$-subgroup $R/S$ of order $49$. Both $T/S$ and $R/S$ are cyclic, and normal.
Then $T$ has order $5\times 29$, so it has unique normal, hence characteristic, subgroup  $T_1$ of order $5$. Since $T_1$ is characteristic in $T$ which is normal in $G$, $T_1$ is normal in $G$. $G$ acts on $T_1$ by conjugation, but the automorphism group of $T_1$ has order $4$ and $|G|$ is odd, so the action is trivial, and $T_1$ is central in $G$.
Let $R/S$ be generated by s coset $aS$ of order $49$. Then $a$ acts on $S$ by conjugation. The order of the automorphism group of $S$ is $28=4\cdot 7$, so $a^7S$ is an element of order $7$ which must act trivially on $S$. Therefore $b=a^{7\cdot 29}$ must have order $7$ in $G$. Also it centralizes $S$. Let $B=\langle b\rangle$. Then the subgroups $T_1, B, S_1$ form a direct product in $G$ (they pairwise commute elementwise, and the intersection of each of them with the product of the other two is trivial by the Lagrange theorem). Therefore $T_1B$ is a central subgroup of $G$ of order $35$ (here we use the fact that $T_1$ is central $b$ commutes with $a$ and all elements of $S$, hence is central in $G$).
