# No group of order $400$ is simple - clarification

I was reading through a proof that no group of order $$400$$ is simple which can be found here: https://math.stackexchange.com/a/79644/169389

Here is an outline for a solution.

First of all, $$|G| = 400 = 2^4 \cdot 5^2\$$. By Sylow's theorem we know that the number of Sylow 5-subgroups must be a divisor of $$2^4$$ and that it is $$1$$ modulo $$5$$. Thus it is either $$1$$ or $$2^4$$. If there is only one Sylow 5-subgroup, it must be normal.

For the other case, suppose first that the intersections of different Sylow 5-subgroups are always trivial. By counting elements you can conclude that $$G$$ has exactly one Sylow 2-subgroup, which is then normal.

If we have Sylow 5-subgroups $$P$$ and $$Q$$ such that $$P \cap Q \neq \{1\}$$, then $$|P \cap Q| = 5$$. Therefore $$P \cap Q$$ is normal in $$P$$ and $$Q$$, and thus is normal in the subgroup $$\langle P, Q \rangle$$ generated by $$P$$ and $$Q$$. Finally, show that either $$\langle P, Q \rangle$$ is normal in $$G$$ or equals $$G$$.

I am trying to get to grips with general problems like this and feel that this is one of the best explanations, but I still feel that this argument goes too quickly and I still need some clarification on a few things.

First problem: In the second paragraph we consider $$n_5$$(number of sylow 5-subgroups)$$=2^4$$ and suppose that the intersections of different Sylow $$5$$-subgroups $$=1$$.

When he says "by counting the elements", I guess he means the intersections of different Sylow $$5$$-subgroups which would be $$1$$ which we have assumed, how can we now conclude that $$n_2=1$$?

Second problem: I am wondering why $$|P \cap Q| = 5$$ and not $$5$$ or $$25$$, I think that it may be because they are different their intersection would have to be $$5$$ since if the intersection is only of order $$25$$ when the groups are the same and of order $$25$$, but if this could be clarified, it would be greatly appreciated

Third problem: To show $$\langle P, Q \rangle$$ is normal in $$G$$ or equals $$G$$. How did a commenter discern that the possible orders for $$\langle P, Q \rangle$$ are $$50,100$$ and $$200$$, why not $$40$$ or $$25$$?

1. What he means by "counting the elements" is this: If there are $2^4 = 16$ subgroups of order $25$, all intersecting trivially, then they alone account for $16\cdot 24= 384$ elements of order $5$ or $25$ in $G$. That means there is only room for $399-384 = 15$ elements of order $2, 4, 8$ or $16$ ($399$ because the identity element must also be made room for). Together with the identity element these elements must be the unique Sylow-$2$-subgroup of order $16$.
2. If $|P\cap Q| = 25$, with $|P| = |Q| = 25$, then yes, $P$ and $Q$ would share every element, and therefore be equal.
3. If $|P\cap Q| = 5$, then $|P\cup Q| = 45$, and we must have $P\cup Q \subseteq \langle P, Q\rangle$, as sets. Therefore, $|\langle P, Q\rangle| \geq 45$, and the only options that statisfy Lagrange's theorem are $50$, $100$ and $200$.
(I) If $P_i, 1 \leq i \leq 2^4$ denote the distinct 5-Sylow subgroup of $G,$ then the number of non-identity element in $\cup_{1 \leq i \leq 2^4}P_i$ is $2^4 \cdot 24 = 384.$ The rest of the non-identity elements must belong to 2-Sylow subgroups and a 2-Sylow subgroup contains 15 non-identity elements. So it must be unique.
(II) $P \cap Q$ is a subgroup of $P$ and $Q.$ Since $|P| = |Q| = 25,$ the only choices for $|P \cap Q|$ is 1, 5, 25 (Lagrange's theorem). By assumption $|P \cap Q| \neq 1$ and $P, Q$ are distinct. So $|P \cap Q| = 5.$
(III) $\langle P, Q \rangle$ is a subgroup of $G$ and also contains both $P$ and $Q.$ So $|\langle P, Q \rangle|$ must divide $|G|$ and also must be divisible by $|P| = |Q|$ (Lagrange's theorem).