# Prove that a group of order 2000 or 4000 is never simple

Suppose that $G$ is a group with order $2000$. Prove that the group is not simple. How about when the group order is $4000$?

Here is my work so far, for the order $2000$ case:

Note that $2000 = 2^45^3$.

Suppose that $G$ is simple. Then all of its Sylow groups are not normal in $G$. Hence by Sylow theorem $n_5 := |Syl_5(G)| = [G:N_G(P_5)] = 1 + 5k \space| 2^4 = 16$ for some $k$ a natural number not zero (here, $P_5$ is a Sylow-$5$ group). It follows that $n_5 = 16$, so that by Lagrange's Theorem $|N_G(P_5)| = 125.$ Let $P$ and $Q$ be distinct Sylow-$5$ groups, and let $I :=P \cap Q$. Notice that the set $PQ:=\{pq \space|\space p \in P, q\in Q\}$ is a subset of $G$ and hence must have an order at most $2000$. We know that $|PQ| = \frac{|P||Q|}{|P \cap Q| }$. Because $I$ is a subgroup of $P$ and $P$ has an order of $125$, by Lagrange's theorem we have that $|I| = 1, 5, 25$ (notice that it cannot be 125 since $P$ and $Q$ are distinct). If $|I| = 1$ or $5$, then $|PQ| \geq 5^6$ or $5^5$, respectively, both of which are greater than $2000$, a contradiction. Hence it must be that $|I| = 25$.

But I wasn't sure how to go about from here. I've already considered $n_2 := |Syl_2(G)|$ as well, but that didn't seem to work either. Maybe I should consider other directions. My professor was hinting that I might have to use the following theorem:

Suppose $G$ is a finite group and let $H$ be a $p$-subgroup of $G$. If $p \space | \space [G:H]$, then $p \space | \space [G: N_G(H)]$.

But I wasn't sure how to use this.

I am thinking that a similar method for solving $2000$ might work for $4000$ as well, but I haven't got the case of $4000$ either.

I also want to prove this without using Burnside's theorem.

• Does Burnside's theorem work for you: both 2000 and 4000 are divisible only by 2 and 5, so they are solvable. Since they are not prime, all groups of those orders are not simple? – Parcly Taxel Dec 9 '17 at 3:59
• Burnside told you that a group whose order is divisible by at most two primes is solvable... – user441558 Dec 9 '17 at 4:00
• 'Suppose that $G$ is simple. Then all of its Sylow groups are not simple.' This is not true... Consider $G=A_5$. – user441558 Dec 9 '17 at 4:02
• 'Suppose $G$ is a finite group and let $H$ be a $p$-subgroup of $G$. If $p\mid [G:H]$, then $p\mid [G: N_G(H)]$.' This is also false. Consider $G=A_4$ and $H=\langle(12)(34)\rangle$. – user441558 Dec 9 '17 at 4:47
• First replace "are not simple" by "are not normal". After that you could continue as follows: $P\cap Q$ being a subgroup of $P$ a nilpotent group must have $N_P(P\cap Q) \neq P\cap Q$ and hence must have $N_P(P\cap Q)=P$. Similarly $N_Q(P\cap Q)=Q$ and hence $|N_G(P\cap Q)| > |PQ|=625$. But then $|N_G(P\cap Q)|$ is either 1000 or 2000. In the first case we have subgroup of index $2$ which must be normal and in the second $P\cap Q$ is a normal subgroup. – Nex Dec 9 '17 at 7:01

## 1 Answer

You got as far as $|I|=25$. Note that $I$ has index $5$ in both $P$ and $Q$, so it is normal in them both. So $N_G(I)$ contains more than one Sylow $5$-subgroup, and hence it contains at least $16$. So $|N_G(I)|$ is divisible by $125$ and by $16$.

So, for $|G|=2000$, $N_G(I) = G$, and $G$ is not simple. For $|G|=4000$, $N_G(I)$ could have index $2$ in $G$, but then again $G$ is not simple.