# If G is not commutative, then is there always a subgroup that is not a normal subgroup?

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!

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This question may be a duplicate of the following one: math.stackexchange.com/questions/37096/…. –  Pete L. Clark Sep 16 '11 at 2:45
@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? –  Zev Chonoles Sep 16 '11 at 6:38
See also this question: Can a non-abelian subgroup be such that the right cosets equal the left cosets?, and especially this answer (by Robin Chapman), and the comments. –  Pierre-Yves Gaillard Sep 16 '11 at 11:40
@Zev: sure, what you suggest sounds quite reasonable. –  Pete L. Clark Sep 16 '11 at 12:55

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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thank you for the nice counterexample! –  user12205 Sep 16 '11 at 11:42