# Proof that no group of order $525$ is simple

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 $$N_G(P \cap P')$$