If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.

By the Sylow theorem, we have that the number of $2$-sylow subgroups of $G$ $n_2$ satisfy that $$ n_2 \equiv1\mod2\mbox{ and } n_2\mid5^6 $$ Similarly for $n_5$ we have, $$ n_5 \equiv1\mod5\mbox{ and } n_5\mid2^4 $$ Hence, $$ n_2 \in \{1,5,5^2,5^3,5^4,5^5,5^6\}, \mbox{ and } n_5 \in \{1,16\} $$

But assuming that none of the $n_p'$s are one and using sylow theorems, I can't surpass the order of $G$ as the professor show us in class with one example. Now I am pretty sure I will need another approach but nothing comes to my mind. Any help would be appreciated.

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    $\begingroup$ One line of argumentation could be that Burnsides p-q-theorem implies that G is solvable, and then under this condition - if G were simple - it would be cyclic, which contradicts the order of G. $\endgroup$ – Teddan the Terran Dec 27 '17 at 22:20
  • $\begingroup$ can I assume that G is simple and get a contradiction ? $\endgroup$ – Richard Clare Dec 27 '17 at 22:28
  • $\begingroup$ Yes. In some versions of Burnside's Theorem, the claim is even that a group of order $p^a q^b$ (where p and q are primes and a and b are positive integers) cannot be simple. But usually the statement is that such a group is solvable. This then implies non-simplicity via a straightforward argument. $\endgroup$ – Teddan the Terran Dec 27 '17 at 22:33

If $n_5=1$, you are done.

If $n_5=16$, take one such 5-subgroup $H$. Its index is $2^4=16$. The index of its normal core $Core (H) $, per well-known Poincare's theorem, divides $16! $, thus it divides $\gcd (2^45^6,16!)=2^45^3$. In other words, $Core (H) $ is of index at most $2^45^3$ so it cannot be trivial - this gives you a nontrivial normal subgroup of $G$.

  • $\begingroup$ Note we are talking about the index here. $|G:Core (H)|=\frac {|G|}{|Core (H)|} $ as both are finite. For example, $|G:Core (H)|=1$ would not mean that $Core (H)$ is trivial; it would mean that $Core (H)=G$ . $\endgroup$ – user491874 Dec 27 '17 at 23:34
  • $\begingroup$ $Core (H ) $ also cannot be the whole of $G $ because $Core (H)\le H $. $\endgroup$ – user491874 Dec 27 '17 at 23:37
  • $\begingroup$ math.stackexchange.com/questions/2582523/… $\endgroup$ – Richard Clare Dec 28 '17 at 1:04
  • $\begingroup$ Thanks, had fun solving that other question, though not much luck in typing the answer ;) $\endgroup$ – user491874 Dec 28 '17 at 1:41
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    $\begingroup$ It is because I thought the opposite - the other answer seemed more succinct. Never mind. $\endgroup$ – user491874 Dec 28 '17 at 1:45

As you state, $n_5$ is either $1$ or $16$. If it's the former, we are done. If it is $16$, then there is a homomorphism from $G\to S_{16}$ given by the fact that $G$ acts on the $5$-sylow subgroups by conjugation. But notice that the prime factorization of $|S_{16}|=16!$ contains exactly three copies of $5$ (coming from $5$, $10$ and $15$), but $|G|=2^45^6$ contains six copies of $5$. So this homomorphism cannot be injective so that it's kernel is a normal subgroup of $G$. So $G$ is not simple.

This expounds the well known aphorism that "Groups, as men, will be known by their actions".

  • $\begingroup$ Do you happen to know whether this kernel is the intersection of all those 5-groups? $\endgroup$ – user491874 Dec 27 '17 at 23:08
  • $\begingroup$ Love it. Thanks. $\endgroup$ – Richard Clare Dec 27 '17 at 23:19
  • $\begingroup$ Actually here I am trying to be lazy and ask if my answer produces the same normal subgroup as your answer - before going off to try to figure it out myself. Thanks anyway! $\endgroup$ – user491874 Dec 27 '17 at 23:29
  • $\begingroup$ @user8734617 You are correct. Just see bullet point #2 in the proof in the page you linked in your answer. Apparently you and I were talking about the same thing. The proof of Poincaré's theorem is pretty much this kind of homomorphism to a symmetric group. $\endgroup$ – Ravi Dec 28 '17 at 0:15

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