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Suppose $G$ is a simple group of order $315=3^2\times 5\times 7.$
Then I can prove that $G$ has $84$ elements of order $5$ and $90$ elements of order $7.$

But when we consider the Sylow $3$ subgroups of $G,$ they may not be distinct.
For any two distinct Sylow $3$ subgroups we have $|H_1\cap H_2|=1$ or $|H_1\cap H_2|=3.$
How can I find the number of elements of order $9$ of $G$ ?

Travis Willse
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Bumblebee
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    I did some googling and found links explaining why it can't be simple. The first one argues by counting the elements: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2007&task=show_msg&msg=3149.0001.0002 and http://cklixx.people.wm.edu/teaching/math430/hwk08-s.pdf – coldnumber Jul 13 '15 at 09:07
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    There would have to be $7$ Sylow $3$-subgroups, so $G \le A_7$, but a Sylow $3$-subgroup of $A_7$ has order $9$ and is not normalized in $A_7$ by an element of order $5$, so we get a contradiction. – Derek Holt Jul 13 '15 at 09:15
  • Thank you ColdNumber and DerekHolt :) If you like to post these comment as answers, I can accept one of them. – Bumblebee Jul 13 '15 at 10:25

1 Answers1

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I'll expand a bit on my comment.

I found two explanations for why no group of order 315 is simple. Both use contradiction.

The one on this link explains it by counting the elements; the trickier part of the counting is that instead of looking for elements of order 9 specifically, it looks for elements whose order is a positive power of 3, and finds the least number of such elements that $G$ can have.

Then using the fact that the index of the normalizer of a Sylow $p$-subgroup is $n_p$ and the counting formula, it finds the number of elements of order 15 in $G$.

The contradiction is that adding all those elements results in more than 315.

The other explanation, on this link, is similar to Derek Holt's argument. Using the Sylow theorems it finds that $G$ has 7 Sylow 3-subgroups, so by the embedding theorem (which states that if a finite non-abelian simple group $G$ has a subgroups of index $n$, then $G$ is isomorphic to a subgroup of $A_n$) $G$ is isomorphic to a subgroup of $A_7$. (We know $G$ is not abelian because if it were, any subgroup would be normal, so, since it has nontrivial subgroups, it wouldn't be simple.)

Like the first link, it notes that the normalizer of a Sylow 5-subgroup has an element of order 15 and gets its contradiction: $A_7$ does not have an element of order 15.

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coldnumber
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