I was having a discussion with a friend of mine about some normal group properties and then came up with the question "if G is not commutative, then is there always a subgroup that is not a normal subgroup?" It's probably more easy to solve this in the following form:$$\forall H \leq G : H \lhd G \Rightarrow \forall a,b \in G : ab=ba$$ My question is, can anybody give a proof, or a counter-example (because I don't think it holds) of this theorem? Thanks!
If G is not commutative, then is there always a subgroup that is not a normal subgroup?
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abstract-algebra
group-theory
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1This question may be a duplicate of the following one: http://math.stackexchange.com/questions/37096/help-on-subgroup-and-normal-subgroup. – 2011-09-16
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0@Pete: Good catch; but I think the question is stated better here, so maybe we should close the other one as a duplicate of this one? – 2011-09-16
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0See also this question: [Can a non-abelian subgroup be such that the right cosets equal the left cosets?](http://math.stackexchange.com/questions/3600/can-a-non-abelian-subgroup-be-such-that-the-right-cosets-equal-the-left-cosets), and especially this [answer](http://math.stackexchange.com/questions/3600/can-a-non-abelian-subgroup-be-such-that-the-right-cosets-equal-the-left-cosets/3601#3601) (by Robin Chapman), and the comments. – 2011-09-16
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0@Zev: sure, what you suggest sounds quite reasonable. – 2011-09-16
1 Answers
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The answer is no. The Quaternion Group provides the smallest counter example.
Another way to write your question is the following: "Does there exist a non Abelian group all whose subgroups are normal." Such counter examples to your above conjecture actually have a specific name, and can be completely classified. These are called Hamiltonian Groups.
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0thank you for the nice counterexample! – 2011-09-16