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Problem: Prove there are no simple groups of order $1,000,000$.

So, I have used Sylow's Theorem to get that the number of Sylow $5$-subgroups is either $1$ (and we're done) or $16$.

I am assuming we have $16$, and let $H$ and $K$ be two of these Sylow $5$-subgroups. I'm trying to show $H \cap K$ is normal in $G$.

So, $|H\cap K| = |H||K|/|HK| > |H||K|/|G| \approx 244.14$

So since $H\cap K<H$, $|H\cap K|$ divides $|H|$. Thus, $|H \cap K| = 625$ or $3125$.

IF $|H\cap K| = 3125$, then $[H \cap K:H] = 5$ which is the smallest prime divisor of $15625$. So $H \cap K$ is normal in $H$ and normal in $K$. This forces $HK$ to be contained in the normalizer of $H \cap K$. Since the normalizer of $H \cap K$ is a subgroup of $G$, then its order must divide $G$. The size of $HK$ is $3125^2$, which is contained in the normalizer. So $3125^2$ must divide $1,000,000$. Contradiction.

So the normalizer of $H \cap K$ is all of $G$, and $H \cap K$ is normal.

I am completely stuck on what to do if $H \cap K$ has order $625$. Any help would be much appreciated.

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    $\begingroup$ Please read the mathjax help It really doesn't take much time at all to make this readable. $\endgroup$ – rschwieb Mar 15 '18 at 16:43
  • $\begingroup$ Welcome to math.se! I edited your post so that the math will be typeset with MathJax. For more on how to typeset math, see this post. $\endgroup$ – André 3000 Mar 15 '18 at 16:50
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    $\begingroup$ Thanks! This is my first post. I didn't know how to format yet. $\endgroup$ – Math Lady Mar 15 '18 at 16:51
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With $16$ Sylow $5$-subgroups, your group $G$ embeds into $S_{16}$ (indeed into $A_{16}$). But $10^6\nmid 16!$.

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  • $\begingroup$ It embeds into S16? I thought it would embed into S64? Since 1,000,000 = 2^6 5^6. What is the mapping? $\endgroup$ – Math Lady Mar 15 '18 at 16:48
  • $\begingroup$ And with S64 we don't get the divisibility problem.... $\endgroup$ – Math Lady Mar 15 '18 at 16:49
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    $\begingroup$ @DeborahJeanWeeks $G$ acts on the set of Sylow $5$-subgroups by conjugation, which induces a homomorphism $G \to S_{16}$. Since $G$ is simple the kernel must be trivial, so the map is an embedding. $\endgroup$ – André 3000 Mar 15 '18 at 16:54
  • $\begingroup$ OOOOOOOOH! Duh. I can't believe I wasted so much time going down this rabbit hole. Thank you! $\endgroup$ – Math Lady Mar 15 '18 at 17:00

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