Are all groups of order $156$ solvable?
Groups of order 156
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$\begingroup$
group-theory
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0Yep. The only non-abelian simple group of order less than or equal to 156 has order 60, which doesn't divide 156. But this is worth proving directly from the Sylow theorems. Is this homework? – 2010-12-14
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0Yep, it is.. I wasnt looking for an answer, more of a definition of solvable and how to tackle it. – 2010-12-14
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0The definition is on Wikipedia (http://en.wikipedia.org/wiki/Solvable_group). Are you familiar with examples where the Sylow theorems are used to classify the groups of order n for some reasonably small n? – 2010-12-14
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0I came across http://ca.answers.yahoo.com/question/index?qid=20091128130200AAPZVBY and understand about 50-60% of it. – 2010-12-14
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0@Qiaochu Yuan : What would be a preferred refrences for sylow theorems? – 2010-12-14
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0@Arjang: the Wikipedia articles seem fine to me. @powerchess: you don't need to deduce the full classification to see this result. It's enough to keep finding normal cyclic subgroups. – 2010-12-14
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0@Qiaochu Yuan: are you suggesting that $156=2^2 \times 3 \times 13$, so we have that there is a p-sylow subgroup of order 13 called H. And thus, we have that G/H is a group of order 12, and since all groups of order 12 are solvable, we are done? – 2010-12-14
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0No, you don't know that H is a normal subgroup so the rest doesn't make sense. – 2010-12-14
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0@powerchess: once you prove that the 13-sylow is normal, that works. – 2010-12-14
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0H is normal because it is a unique sylow subgroup (by 3rd Sylow Theorem) and thus is invariant under conjugation (2nd Sylow Theorem). – 2010-12-14
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0@Joe: Thanks, I think I kind of get it, with this logic, why is a group of order 60 not solvable? $60=2^2\times3\times5$. So let the p-sylow group of G of order 5 be H. Then by the same logic, G/H is normal in G. – 2010-12-14
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0@Joe The whole point of these hints is to give powerchess a chance to think about the exercise and _not_ to solve it for him. – 2010-12-14
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0@powerchess You should review the definitions of subgroup, quotient group and normality. $G/H$ is not even a subgroup of $G$, so the sentence "$G/H$ is normal in $G$" doesn't make sense. Further, what exactly tells you that the Sylow 5-subgroup in your example is normal (if indeed that's what you meant)? – 2010-12-14
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0That was an honest mistake, I did not realize I wrote that.. oops.. I think that since 12 does not divide 5, then let p=5, m=12, then n_p=1 (using wikipedia's notation) so it must be normal. – 2010-12-14
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0@powerchess: this is false; actually we can have n_p = 6. You should review the statement of the Sylow theorems more closely. – 2010-12-14
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0Coming to reconsider what I had said, I understand why n_p may be 6. But how do we know which it is? (or can it be both) – 2010-12-14
1 Answers
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Ask GAP:
gap> AllSmallGroups(156, IsSolvable, false);
[ ]
gap>
This gives the list of all groups of order 156 which are not solvable.