# No group of order 400 is simple

I've been given the question of showing that no group of order $400$ is simple. I've tried to attack it via the Sylow theorems for about a week now, but all the tricks and methods I know seem to be failing horribly.

Things I've tried:

Trying to produce a contradiction by giving a map into $S_n$ by elements acting by conjugation on Sylow 5-subgroups doesn't work, since there are 16 such Sylow 5-subgroups, and 400 divides $16!$, so it might very well be an injection and therefore we can't obviously find a nontrivial kernel.

Trying element counting is messy and I can't get it to come out the way I want- for instance, we can show that each of the Sylow 5-subgroups is isomorphic to $\mathbb{Z}_5\times \mathbb{Z}_5$, so there should be at least 125 elements of order divisible by only 5, but I can't see this producing a contradiction with any of the things I can find out about Sylow 2-subgroups.

Anyways, I'm probably missing something fairly obvious, and I would appreciate any hints, solutions, or other help that you could give.

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– Rasmus Nov 6 '11 at 21:20
I'm sorry, I'm not familiar with representation theory yet. I'm only a college sophomore, and this is my first abstract algebra course. This does look like a very nice and useful theorem, but I think I might have some difficulty proving it. – EWEBEWE Nov 6 '11 at 21:28
try the GAP SYSTEM . it's free – Jawad Nov 25 '11 at 20:32

## 1 Answer

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 $G$.

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Thank you very much. – EWEBEWE Nov 6 '11 at 22:23
Dear m.k. sorry for bringing out a thread from 2 years ago, but could you elaborate on the last sentence: $\langle P, Q\rangle$ is normal in $G$ or equals $G$. I can see that possible orders of $\langle P, Q\rangle$ is 50, 100, or 200. When $\langle P, Q\rangle=200$, then it must be normal, because it has index 2. But I don't follow what happens when $|\langle P, Q\rangle|$ is 50 or 100. – Prism Aug 1 '13 at 4:08
@Prism: Here $\langle P, Q \rangle$ contains the set $PQ$ which has order $5^3 = 125$. So $\langle P, Q \rangle$ would be too small in those cases. – Mikko Korhonen Aug 1 '13 at 4:45
@m.k. That's beautiful. Thanks :) and (+1)! – Prism Aug 1 '13 at 4:59