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This is the third part of a set of problems, of which I have solved 2.

I have shown that if $p$ is prime, the group $Aut(\mathbb Z_p)$ is of order $p-1$.

I have shown that $Aut(\mathbb Z_{17})$, $Aut(\mathbb Z_{257})$, $Aut(\mathbb Z_{65537})$ are 2-groups.

The third problem is the one stated in the title; showing that any group of order $286331153$ is abelian. The problem gives the hint that $286331153 = 17\cdot257\cdot65537$.

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Do you remember the Sylow theorems? –  Daniel Fischer May 15 at 15:18
    
Ah, yes. I should have mentioned that. I just can't seem to piece it into this problem. –  Aleksander May 15 at 15:25

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up vote 6 down vote accepted

Your group has order $pqr$, for $p<q<r$ primes, and $pq <r$.

Let $G$ be a finite group of order $pqr$ with $pq<r$ primes. Then $G$ has a normal subgroup of order $r$: we know that $n_r=1+kr\mid pq$. But if $k\geqslant 1$, $n_r=1+kr>1+kpq\not\mid pq$, so $n_r=1$. Thus, we have a cyclic normal $C$ subgroup of order $r$. Of course this has trivial intersection with the other Sylow groups. Thus $G$ is a semidirect product $C\rtimes H$ with $H$ a group of order $pq$.

Now recall that if $H$ is a group of order $p<q$ primes and $p\not\mid q-1$, then $H\simeq C_{qp}$. Since $17\not\mid 256$, we now have $H\simeq C_{qp}$, so $$G=C_r\rtimes C_{qp}$$

You now need to discard the possibility that the semidirect product is nontrivial.

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There are non-abelian groups of order e.g. $2\cdot3\cdot7$, the semi-direct product $C_7\rtimes C_6$. If $a$ is of order $7$, $b$ of order six, and $bab^{-1}=a^3$, then that's what you get. –  Jyrki Lahtonen May 15 at 15:22
    
@JyrkiLahtonen Yes, but $6\mid 7-1$ in that case. –  Pedro Tamaroff May 15 at 15:23
    
True :-) ${}{}{}$ –  Jyrki Lahtonen May 15 at 15:24
    
@JyrkiLahtonen You're right, however. I should be a bit more explicit. –  Pedro Tamaroff May 15 at 15:24

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