# If the Order is an Even Perfect Number, then a Group is not Simple ## Problem 74

(a) Show that if a group $G$ has the following order, then it is not simple.

1. $28$
2. $496$
3. $8128$

(b) Show that if the order of a group $G$ is equal to an even perfect number then the group is not simple. Add to solve later

## Hint.

Use Sylow’s theorem.
(See the post Sylow’s Theorem (summary) to review the theorem.)

## Proof.

### (a) A group of the following order is not simple

#### (1) A group of order $28$

Note that $28=2^2\cdot 7$. The number $n_7$ of the Sylow $7$-subgroups of $G$ satisfies
$n_7 \equiv 1 \pmod{7} \text{ and } n_7|2^2.$ Thus, the only possible value is $n_7=1$. The unique Sylow $7$-subgroup is a proper nontrivial normal subgroup of $G$, hence $G$ is not simple.

#### (2) A group of order $496$

Note that $496=2^4\cdot 31$. By the same argument as in (1), there is a normal Sylow $31$-subgroup in $G$, hence $G$ is not simple.

#### (3) A group of order $8128$

We have $8128=2^6\cdot 127$, where $127$ is a prime number.
Again the same reasoning proves that the group $G$ has the unique normal Sylow $127$-subgroup in $G$, hence $G$ is not simple.

### (b) If the order is an even perfect number, a group is not simple

From elementary number theory, all even perfect numbers are of the form
$2^{p-1}(2^p-1),$ where $p$ is a prime number and $2^p-1$ is also a prime number.
(For a proof, see the post “Even Perfect Numbers and Mersenne Prime Numbers“.)

Suppose the order of a group $G$ is $2^{p-1}(2^p-1)$, with prime $p$, $2^p-1$.
Then the number $n_{2^p-1}$ of Sylow $(2^p-1)$-subgroup satisfies
$n_{2^p-1}\equiv 1 \pmod {2^p-1} \text{ and } n_{2^p-1}|2^{p-1}.$ These force that $n_{2^p-1}=1$.
Therefore the group $G$ contains the unique normal Sylow $(2^p-1)$-subgroup, hence $G$ is not simple.

### Similar problem

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

## Comment.

In about 300 BC Euclid showed that a number of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime numbers.
(If $2^p-1$ is a prime number, then $p$ must be a prime.)
The converse was proved by Euler in the 18th century. Namely, Euler proved that any even perfect number is of the form $2^{p-1}(2^p-1)$ with prime $2^p-1$.

The number of the form $2^p-1$ is called a Mersenne number and if it is a prime number, then it is called a Mersenne Prime.
It is unknown whether there are infinitely many Mersenne prime numbers.

Also it is unknown whether there is an odd perfect number. Add to solve later

### 2 Responses

1. 08/21/2016

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