Are Groups of Order 100, 200 Simple?

Group Theory Problems and Solutions in Mathematics

Problem 100

Determine whether a group $G$ of the following order is simple or not.

(a) $|G|=100$.
(b) $|G|=200$.
 
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Hint.

Use Sylow’s theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow’s theorem.

Proof.

(a) When $|G|=100$.

The prime factorization of $100$ is $2^2\cdot 5^2$. Let us determine the number $n_5$ of $5$-Sylow subgroup of $G$.
By Sylow’s theorem, we know that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^2$.
The only number satisfies both constraints is $n_5=1$. Thus there is only one $5$-Sylow subgroup of $G$. This implies that the $5$-Sylow subgroup is a normal subgroup of $G$.
Since the order of the $5$-Sylow subgroup is $25$, it is a proper normal subgroup. Thus, the group $G$ is not simple.

(b) When $|G|=200$

The prime factorization is $200=2^3\cdot 5^2$.
We again consider the number $n_5$ of $5$-Sylow subgroups of $G$.

Sylow’s theorem implies that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^3$.
These two constraints has only one solution $n_5=1$.
Thus the group $G$ has a unique proper normal $5$-Sylow subgroup of order $25$. Hence $G$ is a simple group.

Similar problem

For an analogous problem, check out the post: If the order is an even perfect number, then a group is not simple
 

Comment.

This is the 100th problems in this blog.
To post 100 problems were not so simple.

The next goal is to archive the 200th problem.
This problem suggests that this goal is again not simple.
(Update: On 11/25/2016 I achieved the 200th problem: Maximize the dimension of the null space of $A-aI$.)


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2 Responses

  1. 12/21/2016

    […] Groups of order 100, 200. Is it simple? […]

  2. 01/07/2017

    […] For an analogous problem, check out: Groups of order 100, 200. Is it simple? […]

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