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March 19th, 2015, 04:48 AM   #1
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how to find group and subgroup

how to find simple group of order 168. what is the number of subgroups of G of order 7?

Thanks in advance
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March 19th, 2015, 04:51 PM   #2
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I am puzzled by this. If, as you appear to be saying, you know nothing about "simple groups", where did you get this problem?

I, frankly, don't know how to prove it, but Wikipedia says "The second smallest nonabelian simple group is the projective special linear group PSL(2,7) of order 168, and it is possible to prove that every simple group of order 168 is isomorphic to PSL"

It is easy to see that the prime factorization of 168 is $\displaystyle 2^2(3)(7)$. Notice that 7 is a prime factor of 168.
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March 19th, 2015, 07:11 PM   #3
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Thanks for your response. I am preparing for national lectureship exam; this was previously asked question from abstract algebra. From your solution, I noticed that prime factor should not be considered (168 = 2 cube (3) (7) ). In that case the answer would be 8 .
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May 10th, 2015, 03:09 AM   #4
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Quote:
Originally Posted by sriims View Post
how to find simple group of order 168. what is the number of subgroups of G of order 7?
By Sylow, the number of subgroups is of the form $1+7k$ and divides $\frac{168}7=24$. Hence there are $1$ or $8$ possible subgroups. If $1$ then this subgroup is normal and so the group is not simple. Hence there are $8$ subgroups of order $7$ in a simple group of order $168$.

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Originally Posted by Country Boy View Post
It is easy to see that the prime factorization of 168 is $\displaystyle 2^2(3)(7)$.
It’s $2^3(3)(7)$.
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