I would like some verification that any group $G$ of order $|G| = 525 = 5^2 \cdot 3 \cdot 7$ is not simple. I've attached my argument below. Please let me know if you see any issues. Thanks
Let $G$ have order $525$ and assume that it is simple. Consider the Sylow $5$-subgroups, each of order $25$. By Sylow's theorems, the number of Sylow $5$-subgroups divides $21$, and is of the form $5k + 1$ for some $k \geq 0$. We assume, of course, that there is not $1$ Sylow $5$-subgroup, so the only possibility is that we have $21$ Sylow $5$-subgroups.
If all these subgroups have a trivial intersection, then we see that there are $24 \cdot 20 + 25 = 505$ distinct elements in these groups, and thus we have $20$ elements remaining. Consider the $7$-subgroups of $G$. There must be $7m + 1$ of them (not $1$, because then we would have a normal subgroup that is non-trivial) and $7m + 1$ must divide 75, thus making our only choice $15$ subgroups of order $7$. These groups all have non-trivial intersection as they are of prime order, and if they are disjoint from the $5$-subgroups they comprise $15*6 = 90$ elements (not counting the identity, which we counted when considering the $5$-subgroups), which is too many for our group. Note that this is the only case, i.e they must be disjoint from the $5$-subgroups because their intersection would have order that divides both $7,25$, thus also $\gcd(7,25)=1$. Thus, we cannot have all the $5$-subgroups with trivial intersection.
Now, we must consider the case when two distinct Sylow $5$ subgroup have intersection with order not equal to $1$. The only possibility is $5$ by Lagrange's theorem (it cannot be 25 as that would mean the two $5$ subgroups coincide). Thus, we have that $|P \cap P'| = 5$. Both $P,P'$ are abelian (they are order $p^2$) so we have $P < N_G( P \cap P'), P' < N_G(P \cap P')$. Applying Lagrange's theorem again, we see that $|N_G(P \cap P')| > |P| = 25$ and $|N_G(P\cap P')|$ divides $525$. The possible orders then are $75,105,175$. If the order is $75$, we see that $[G : N_G(P \cap P')]$ is 7, but $|G|$ does not divide $7!$, so there cannot be an injective homomorphism from $G$ into $S_7$, so we have a homomorphism from $G$ to $S_7$ has a nontrivial kernel, which ends up being a normal subgroup of $G$ that is nontrivial, meaning $G$ is not simple. The same argument can be repeated with the other orders of $N_G(P \cap P')$