Question: Prove that no group of order 10,000 is simple.

Attempt at solution:

By the Sylow Theorems, the number of Sylow 5-subgroups can only be 1 or 16 (as this number needs to divide 10,000 and also be equivalent to 1 mod 5).

If there is 1 Sylow 5-subgroup, then it is automatically a normal subgroup (since its conjugate can only be itself), so the group is not simple.

If there are 16 5-subgroups...

Any help would be appreciated. Thank you!


2 Answers 2


As Serkan mentioned, this is a classic Sylow simplicity problem where the order is too large to be a subgroup of the associated symmetric group. I was expecting to see questions about simple groups of order 2012, and so was presently surprised to see this question from the future. Greetings from the past to our friends from the Y10K! However, I'm sad to see the techniques have not changed over the next 8000 years.

Serkan's proof works nicely for 10000, but 12 years ago we were not so futuristic and had to make do with the following proposition which I hope is useful to you:

Proposition: A group of order $2^4 \cdot 5^n$ for $n \geq 2$ has a normal subgroup of order $5^n$ or $5^{n-1}$.

Proof: Let $G$ be such a group, $P$ a Sylow 5-subgroup. Sylow's theorems show that $N_G(P)$ is either $G$ or $P$, and the former is one of the conclusions. Hence we may assume $N_G(P)=P$. Consider the action of $P$ on the $G$-conjugates of $P$. $P$ fixes $P$, but cannot fix any other conjugate. Writing $16-1$ as a sum of divisors of $5^n$ can only be done as $5+5+5$. Hence in all three orbits $Q=N_P(P^g)=P \cap P^g$ has index 5 in $P$. Since $Q$ is maximal in both $P$ and $P^g$, $Q$ is normal in both $P$ and $P^g$. Hence $N_G(Q)$ is a group of order $2^i 5^n$ with more than one Sylow 5-subgroup. If $i <4$, this contradicts Sylow's theorem, so $i=4$ and $N_G(Q) = G$. $\square$

In fact, we can even be quite explicit about the structure of $G/Q$ in case the Sylow 5-subgroup is not normal. Then we have $G/Q \leq \operatorname{AGL}(1,16)$ has order $2^4 \cdot 5$ and is the unique non-abelian semi-direct product of the elementary abelian group of order 16 with a cyclic group of order 5. Of course $Q$ itself can be any group of order 125. The proof is merely that no other group of order 16 has an automorphism of order 5 (by Burnside's basis theorem), so every other Sylow 2-subgroup would centralize the Sylow 5-subgroup and thus normalize it.

  • $\begingroup$ Thank you Jack :-) Good to know the machinery was in place in year 2000! $\endgroup$
    – Conan Wong
    Aug 29, 2012 at 4:54

$$|G|=10,000=2^4\cdot5^4$$ By Sylow theorems, there are either one or $\,2^4=16\,$ Sylow $\,5\,$-subgroups. If the latter is true, defining $\,P=\,$ Sylow $\,5\,$-sbgp. , we get $$[G:N_G(P)]=16$$ Making act $\,G\,$ regularly (left shift, say) on the set of cosets of $\,N_G(P)\,$, we get a homom. $\,\phi:G\to S_{16}\,$ , which must be injective if we assume $\,G\,$ is simple, and thus our group is isomorphic with a subgroup of $\,S_{16}\,$, which is impossible (why? Lagrange's Theorem)

  • $\begingroup$ I realize this is a bit of an old question, but I was wondering 2 things. 1) Why must $\phi$ be injective? Couldn't $\ker(\phi)=G$? 2) Why does $\phi$ being injective tell us that $G$ is isomorphic with a subgroup of $S_{16}$? Thank you! $\endgroup$
    – lollab
    Oct 25, 2020 at 9:13
  • $\begingroup$ @lollab If a group is simple, then any hjomomorphism from it to any group is either trivial or injective, as the group cannot have non-trivial normal subgroups...In the above case it cannot be $\;\ker G=G\;$ since the the homomorphism determined by the regular representation of $\;G\;$ in $\;S_n\;$ is never trivial. And if $\;\phi: G\to H\;$ is injective then $\;\phi(G)\;$ is a subgroup of $\;H\;$ which is trivially isomorphic to $\;G\;$ as $\;\ker\phi=\{1\}\;$ ... $\endgroup$
    – DonAntonio
    Oct 25, 2020 at 9:33

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