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Question
Show that there are no simple groups of order
, where p is a prime, r is a positive integer, and m < p.
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Solution B
Answered 11 months ago
Step 1
1 of 4Given:
* $p$ is a prime number
* $r$ is a positive integer
* $m<p$
In this exercise, we prove that there can't exist a simple group of order $p^rm$.
*What is a simple group? Which theorem can help us derive whether a group is simple?*
Answered 11 months ago
Step 1
1 of 2Let $p$ be a prime, $r$ a positive integer and $m<p$. By a theorem in the textbook, the number of Sylow p-subgroups of a group of order $p^rm$ is congruent to 1 modulo $p$ and divides $m$, but the only such number is 1. Therefore a group of order $p^rm$ has one Sylow p-subgroup which must be normal by a previous exercise, and therefore it can't be simple.
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