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We know that a subgroup N of an abelian group G must be normal. However, is the reverse necessarily true? An illustration would enlighten me.

Any resource links or names of illustrative texts welcome, as I am new to Group Theory.

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up vote 6 down vote accepted

It is not even true that if every subgroup of $G$ is normal then $G$ must be abelian. The smallest example is the quaternion group of order $8$, $Q_8 = \{\pm 1,\pm i,\pm j,\pm k\}$. The subgroup of order $4$ is normal because it has index $2$; the only subgroup of order $2$ is $\{\pm 1\}$, which is normal because it equals the center of the group.

Groups in which every subgroup is normal are sometimes called "Dedekind groups", with nonabelian Dedekind groups being called "Hamiltonian groups".

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I would like to add to this, since I was intrigued by this problem today. The structure of Hamiltonian groups was described by Baer as follows: A group $G$ is Hamiltonian iff $G\cong C\oplus T\oplus Q_8$ where $C$ is a direct sum of cyclic groups of order 2, $T$ is a torsion subgroup with only odd torsion. So every Hamiltonian group contains $Q_8$ in a very fundamental way, in that the quotient $G/Q_8$ is abelian (where $Q_8$ is embedded in the natural way) –  you Oct 19 '12 at 11:57

By the converse, I assume you mean if a subgroup $N$ of a group $G$ is normal, must $G$ be abelian? If so, this is definitely not true, and is a reason why normal subgroups are interesting and important. The simplest example of a normal subgroup in a non-abelian group is the subgroup $N = \langle (1 2 3) \rangle \subset S_3$.

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Or perhaps the question is whether abelian groups are the only ones in which every subgroup is normal, in which case the answer is also no, and the examples showing it have a Wikipedia article: en.wikipedia.org/wiki/Hamiltonian_group –  Jonas Meyer May 5 '11 at 4:22
    
Thanks to you all of you –  Bhaskar Dey May 5 '11 at 4:33

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