7
$\begingroup$

Is there a neat way to show that a group of order 160 is not simple without directly quoting Poincare's theorem?

I am thinking of maybe using the Sylow theorems to say that in order that the group is not simple we must have $n_2=5,\,\,\,n_5=2^4$ (due to the Sylow constraints). So then there are $2^4\times 4$ order 5 elements. But then I can't say that there are $5\times 2^5$ elements with not-equal-to-5 orders since these subgroups may intersect because they don't have prime orders. So what can we do next, or is this not even the right direction to start?

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    Is there any reason why you would like to avoid Poincare?2012-01-27
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    @KannappanSampath: I have nothing against Poincare per se, but I was told that there is not-too-convoluted way to solve it... And maybe the alternate way would be illuminating in some way! ;)2012-01-27
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    You can get the number of even-order elements up to 15 + 4*8, since each sylow-2 subgroup must at least be different from the others in the 8 elements with order relatively prime to the group order, but that still only gets you to a total of 112 elements accounted for.2012-01-27
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    Counting arguments will go through. Look at the Normalizer of the intersection of two Sylow 2-Subgroups. (Also note that you cannot have more than one distinct Sylow 2-Subgroup.)2012-01-27
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    @KannappanSampath: I still don't quite see it, would you mind expanding a bit? Thanks.2012-01-27
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    You could also use Burnside's pq-theorem (stating that a group of order divisible only by two distinct primes can not be simple) but this might be a bit of an overkill for this problem.2012-01-27
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    @SebastianSchoennenbeck: Thanks, actually yours is a very nice suggestion!2012-01-27
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    Another way I just thought of: Count the 2-Sylow-subgroups. There is either a single one (which consequently would be normal) or 5 distinct ones. In the later case the group acts transitively on 5 points and we get a non-trivial homomorphism into $S_5$ which is a group of order $120$. Therefore the kernel has order greater than $1$ and is a non-trivial normal subgroup.2012-01-27
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    @SebastianSchoennenbeck: Thanks, again! :)2012-01-28

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