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Prove the order of finite simple group is not perfect and even number.

From elementary number theory, all even perfect numbers are of the form $$ 2^{p-1}\left(2^{p}-1\right) $$ where $p$ is a prime number and $2^{p}-1$ is also a prime- number.

Suppose the order of a group $G$ is $2^{p-1}\left(2^{p}-1\right),$ with prime $p, 2^{p}-1$ Then the number $n_{2^{p}-1}$ of Sylow $\left(2^{p}-1\right)$ -subgroup satisfies $n_{2^{p}-1} \equiv 1 \quad\left(\bmod 2^{p}-1\right)$ 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 $\left(2^{p}-1\right)$ -subgroup, hence $G$ is not simple. But is this proof unique ? I want proof without use sylow subgroup.

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    $\begingroup$ Please provide some context. Where did you encounter this problem? What are your thoughts on the problem? What have you tried so far? $\endgroup$ Mar 22, 2020 at 13:58
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    $\begingroup$ Are you allowed to use Cauchy's Theorem, which says that if the prime $q$ divides $|G|$ then $G$ has an element of order $q$? If so, let $R$ be the subgroup of order $q:=2^p-1$ generated by such an element, let $g \in G$ and let $S=g^{-1}Rg$. Then we must have $g^{-1}Rg=R$ (and hence $R$ is normal in $G$), because otherwise the order of $|RS|$ would be $q^2$, which is greater than $|G|$. $\endgroup$
    – Derek Holt
    Mar 22, 2020 at 15:18

1 Answer 1

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Any even perfect number is of the form $2^{n-1} \cdot (2^n-1)$ for some $n$ such that $2^n-1$ is prime. By Sylow's Theorems, the number of $(2^n-1)$-Sylow groups divides $2^{n-1}$ and is congruent to $1$ modulo $2^n-1$. This implies that the number of $(2^n-1)$-Sylows is $1$, and hence the $(2^n-1)$-Sylow is normal.

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    $\begingroup$ But four minutes after you posted this good solution to the question the OP added your good proof to the question without any acknowledgement and said they didn't want to use Sylow. Makes on wonder why one helps, doesn't it? $\endgroup$ Mar 22, 2020 at 14:46

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