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I would like to know the answer to following question: Is it true that if $G$ is a group such that $N\triangleleft M$ and $M\triangleleft G$ then $N\triangleleft G$, for all subgroups $N,M$ of $G$, then $G$ is an abelian group?

I would appreciate any help.

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No: there are groups in which every subgroup is normal but the group is not abelian (known as Dedekind groups). $Q_8$ is an example. Such a group satisfies your condition. –  Arturo Magidin Feb 1 '12 at 20:40
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up vote 8 down vote accepted

Groups with the property that $N\triangleleft M\triangleleft G$ implies $N\triangleleft G$ are sometimes called $t$-groups (or $T$-groups). (Added: More recent literature with access to a wider variety of typeface seems to use $\mathfrak{T}$-group).

For two trivial examples of $t$-groups that are not abelian:

  1. Any nonabelian simple group is a $t$-group, since we must have $M=G$ or $M=\{e\}$.
  2. Any Dedekind group (groups in which every subgroup is normal) is a $t$-group. The quaternion group of order $8$ is an example of a nonabelian Dedekind group.

The name "$t$-group" seems to have been introduced by Olga Taussky and Ernest Best, in A class of groups, Proc. Roy. Irish Acad. Sect. A 47 (1942) 52-62, MR0006539 (4,2b). They proved that a finite $p$-group is a $t$-group if and only if it is abelian or a Dedekind group, and that a group is a $t$-group if it can be generated by elements $p_1,\ldots,p_n,q$ such that $p_i^m=q^n=1$, $p_ip_j = p_jp_i$, and $p_iq = qp_i^r$ where $\gcd(n,m)=1$, and $rn\equiv 1 \pmod{m}$; in particular, groups with all $p$-Sylow subgroups cyclic are $t$-groups.

Giovanni Zacher (Caratterizzazione die $t$-gruppi finiti risolubili, Ricerche Mat. 1 (1952), 287-294, MR0053104 (14,722b)) studied the finite soluble $t$-groups, and shows that a finite soluble group $G$ of order $p_1^{a_1}\cdots p_k^{a_k}$, $p_1\gt\cdots\gt p_k$, is a $t$-group if and only if (i) all Sylow subgroups are abelian or Dedekind groups; and (ii) $G$ has a sequence $S_1,S_2,\ldots,S_k$ of Sylow subgroups, $S_i$ of order $p_i^{a_i}$, such that every $s_i\in S_i$ conjugates elements of $S_j$, $j\lt i$, into powers of themselves. (In particular, this would hold for groups in which all Sylow subgroups are cyclic).

Wolfgang Gaschütz gave an alternative characterization in Gruppen, in denen das Normalteilersein transitiv ist, J. Reine Angew. Math. 198 (1957), 87-92, MR0091277 (19,940b): If $G$ if finite and solvable, then $G$ is a $t$-group if and only if there exists $L\triangleleft G$ such that $L$ is abelian of odd order, $\gcd(|L|,|G/L|)=1$, $G/L$ is a Dedekind group, and conjugation induces power automorphisms in $L$. (In particular, a subgroup of a finite solvable $t$-group is necessarily a $t$-group).

Derek J.S. Robinson studied the infinite solvable case in Groups in which normality is a transitive relation, Proc. Cambridge Philos. Soc. 60 (1964), 21-38, MR0159885 (28 #3101). He proved that a solvable $t$-group is necessarily metabelian (there is an abelian normal subgroup $M$ such that $G/M$ is also abelian); that finitely generated solvable $t$-groups are either finite or abelian; and that if $G$ is a torsion $t$-group and $C$ is the centralizer of $[G,G]$, then $[G:C]$ countable implies that $G$ splits over $[G,G,G]$.

A few years later, in A note on finite groups in which normality is transitive, Proc. Amer. Math. Soc. 19 (1968), 933-937, MR0230808 (37 #6366), Robinson proved that the following are equivalent for finite groups: (i) $G$ is a solvable $t$-group; (ii) Every subgroup of $G$ is a $t$-group; (iii) Every subgroup of a $p$-Sylow subgroup $P$ of $G$ is normal in $N_G(P)$ for all primes $p$.

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$t$-group is a reaaaaally bad name. –  Mariano Suárez-Alvarez Feb 1 '12 at 21:37
    
@MarianoSuárez-Alvarez: Agreed: blame Olga Taussky! (The more recent literature seems to use $\mathfrak{T}$-group...) –  Arturo Magidin Feb 1 '12 at 21:38
    
+1, I appreciate your literature studies as much as your expository work :) –  Myself Feb 1 '12 at 21:45
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