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Show whether or not $G$ is simple If $|G| = 1280$ and explain your reasoning.

I am told I have to use Sylow's theorems.

I have begun by finding that $n_{2}\in \{1,5\}$ and $n_5\in \{1,256\}$. I then know that if $n_2=1$ or $n_5=1$, then $G$ would not be simple. So I am trying to see if $n_2=5$ AND $n_5=256$ but am struggling with how to do this part.

Can I say that if $G$ is simple, $|G| =1280$ must divide 5!, which is not the case?

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    It's not really relevant, but how did you exclude the possibility that $n_5=16$? – 2017-02-13

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Sylow's theorems say that Sylow $p$-subgroups are conjugate to each other. This gives a group action of $G$ on the set of Sylow $p$-subgroups which is transitive. In other words you get a group morphism from $G$ to $S_{n_p}$ which is not the trivial map as the action is transitive. What are the possibilities for the kernel of this map if $G$ were simple?

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    If $G$ were simple then would the kernel have to be trivial? So couldn't contain more than one element? I struggling to get my head round the link – 2017-02-11
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    @harry55 Yes. So you would have an injection from $G$ to $S_{n_p}$ ($G$ is isomorphic to a subgroup of $S_{n_p}$), so what does that tell you about $|G|$? (I am trying to lead you to the proof of your last statement in your original question which does answer the problem) – 2017-02-11
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    That there's an injection from $G$ to $S_5$ so |G| would have to equal 120 if it were simple as isomorphic groups have the same order? – 2017-02-11
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    @harry55 It doesn't have to be an isomorphism. Only a one to one map. But the image is a subgroup (and the map is an isomorphism onto its image) and by Lagrange's the size of the image has to divide $|S_{n_p}|=n_p!$ (which is 120 in your case). – 2017-02-11
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The cyclic group $\mathbf Z/1280\mathbf Z$ is a counter-example, as as an abelian group is simple if and only if it has prime order.

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    Though the question isn't very clear, I bet that he wanted to show that all groups with that prescribed order are not simple (using the usual first year algebra tricks). – 2017-02-10
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    For this particular question I can't give a counter example, I must use Sylow's theorems and the facts that if $G$ is simple then $|G|$ divides $n!$, where $H$ is a subgroup of index >1 in $G$. So can I say that as 1280 does not divide 5!, the group cannot be simple? – 2017-02-11