It's subgroup not "supgroup".
Suppose that $G$ has a normal Sylow $5$-subgroup $N$. Then $|G/N|=21$ and $G/N$ has a normal Sylow $7$-subgroup $M/N$. Then $|M|=35$, so $M$ has a normal Sylow $7$-subgroup $P$. Since $M \le N_G(P)$ with $|G:M|=3$, $|G:N_G(P)|=1$ or $3$. But we cannot have $n_7=3$, so $N_G(P)=G$ and $G$ has a normal Sylow $7$-subgroup.
Alternatively, $P$ is a characteristic subgroup opf $M$ which is normal in $G$, so $P$ is normal in $G$.
The proof in the other case is similar.
Note that Sylow theorems give you necessary conditions for the number of Sylow $p$-groups. Those conditions are not sufficient. In your case, you have proven that $|Syl_5(G)|=1$ and $|Syl_3(G)|\in\{1,25\}$. This means that $|Syl_3(G)|$ and $|Syl_5(G)|$ cannot take any other values than those you've noted - but it doesn't mean they must take all of them.
(The case where those numbers are all $1$ is a bit of an exception. There is always a group where all the numbers of Sylow $p$-groups are $1$. Indeed - take a direct sum of the group's Sylow $p$-groups!)
Which leaves us with the case $|Syl_5(G)|=1$ and $|Syl_3(G)|=25$, for which we still don't know whether it is possible or impossible. Suppose that it is possible, and let's try to either derive a contradiction or construct the group $G$ in which this is true. What we do know is that the Sylow $5$-group is unique (call it $H$, $|H|=25$) and is therefore normal in $G$. On the other hand, you can pick a Sylow $3$-subgroup $K$, $|K|=3$, which is most certainly not normal (Sylow's 2nd theorem - Sylow $3$-groups are all conjugates of each other, so $K$ has $25$ conjugates). We also know that:
Now, the map $\theta_a:h\mapsto a^{-1}ha$ is an automorphism of $H$ and its order must divide the order of $a$ - so its order is either $1$ or $3$. If its order is $1$, however, it means that $a^{-1}ha=h$, i.e. $ah=ha$ for all $h\in H$. This means that every element of $H$ commutes with every element of $K$, and so $G$ turns up a direct sum of $H$ and $K$ - and so $Syl_3(G)=1$.
Thus, the question here really boils down to: is there an automorphism of order $3$ of either $C_{25}$ or $C_5\times C_5$?
- If there is such an automorphism $\theta$, then you can use it to construct a semidirect product $H\rtimes_\theta K$, with $75$ elements, in which $K$ will be most definitely not normal subgroup - therefore $|Syl_3(G)|$ will be $25$.
- If there is no such automorphism, then $\theta_a$ must be trivial and we are back to the case $|Syl_3(G)|=1$
Let's check the two cases:
- $H\cong C_{25}$: Let $b$ be a generator of $H$. An automorphism of $H$ sends $b$ into another generator $b^i$ (where $i$ is coprime to $25$). The order of this automorphism is the smallest number $n$ such that $i^n\equiv 1\pmod{25}$, for which Euler's theorem tells us $n\mid 20$ (as $\varphi(25)=20$). As $3\not\mid 20$, this automorphism cannot be of order $3$.
- $H\cong C_5\times C_5$: One can see that $H$ is then a vector space over $\mathbb Z_5$ of dimension $2$ and every automorphism of it is given by an invertible matrix $A=\begin{bmatrix}p&q\\r&s\end{bmatrix}\in M_2(\mathbb Z_5)$. So we are looking for the $2\times 2$ matrix $A\in M_2(\mathbb Z_5)$ such that $A\ne I$ but $A^3=I$. As it happens, there is such a matrix: take, for example, $A=\begin{bmatrix}-1&1\\-1&0\end{bmatrix}$.
Let us just show how this automorphism acts on $C_5\times C_5$: if the elements of $C_5\times C_5$ are represented as $u^iv^j$ where $u$ and $v$ are the generators of the two $C_5$'s and $i,j\in\mathbb Z_5$, then $\theta_A(u^iv^j)=u^{-i+j}v^{-i}$, because $\begin{bmatrix}-1&1\\-1&0\end{bmatrix}\begin{bmatrix}i\\j\end{bmatrix}=\begin{bmatrix}-i+j\\-i\end{bmatrix}$.
With that automorphism, your group $G$ can be constructed, as shown above, as $(C_5\times C_5)\rtimes_{\theta_A}C_3$, and it will have $25$ Sylow $3$-groups.
Best Answer
First note that since $P$ or $Q$ is normal, $PQ \leq G$. We have $|P|$ and $|Q|$ are coprime, so they must have trivial intersection. Thus $|PQ| = \frac{|P||Q|}{|P \cap Q|} = \frac{15}{1}$. Thus $PQ$ is normal in $G$ since it has index $2$. If $Q$ is normal in $G$ then $P$ has index $3$ in $PQ$. Since $3$ is the smallest prime dividing $|PQ|$, $P$ is necessarily normal in $PQ$ and thus characteristic. If $P$ is normal in $G$ then consider the number of possible subgroups of order $3$ in $PQ$. We know from the Sylow Theorems that it must be congruent to 1 modulo three. Further it must divide 5. The only divisor of $5$ congruent to 1 modulo 3 is 1. Thus there is one Sylow 3 subgroup of $PQ$, namely $Q$.