## Related questions with answers

Question

Show that there are no simple groups of order

$p^rm$

, where p is a prime, r is a positive integer, and m < p.

Solutions

VerifiedSolution A

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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