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How to prove that the group of order 28 which have normal subgroup of order 4 is Abelian group?

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Let $G$ be the group of order $28$, $N$ the normal subgroup of order $4$, and let $K = G/N$ be the quotient group, which has order $7$ and is therefore cyclic. By the Schur-Zassenhaus Theorem, $G$ is a semidirect product of $N$ and $K$. It will be abelian iff the semidirect product is actually direct.

In order to give a nontrivial semidirect product here, it is necessary and sufficient to find a nontrivial homomorphism from $K$ into the automorphism group $\operatorname{Aut}(N)$ of $N$, or equivalently an element of order $7$ in $\operatorname{Aut}(N)$. But $N$ is either cyclic of order $4$ -- in which case $\operatorname{Aut}(N)$ has order $2$ -- or is the Klein group $C_2 \times C_2$, in which case $\operatorname{Aut}(N)$ has order $6$. Either way there is no element of order $7$.

Alternately, consider what Sylow theory has to say about the Sylow $7$-subgroups.

Note: of course the second method, which is only hinted at, is probably what is intended. But the two methods interact in an interesting way with regard to the following more general problem.

Suppose $G$ is a finite group of order $np$ with $p$ a prime number which does not divide $n$, and that there is an abelian normal subgroup $N$ of order $n$. Under what conditions on $N$ and $p$ is $G$ necessarily abelian?

If you use Sylow theory, you get that it is enough for $p$ to be sufficiently large compared to $n$. (One can be explicit; I don't want to be yet to give the OP some time to think about his particular problem.) Now comparing to the approach of showing that any semidirect product of $N$ with the cyclic group of order $p$ is necessarily direct one gets information about the largest prime $p$ which divides the order of the automorphism group $\operatorname{Aut}(N)$. It could be fun to explore this further. (More fun than proving something about groups of order $28$, anyway. This type of overly specific question tends to bore me.)

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Thank you very much! – Aldo Dec 21 '10 at 15:05

This was an answer to a merged question.

No, any group of order $28$ which contains a normal subgroup of order $4$ is necessarily abelian.

The number of $7$-Sylow subgroups must be a divisor of $4$, and also congruent to $1$ modulo $7$, hence there is a unique $7$-Sylow subgroup which is necessarily normal. If $G$ contains a normal subgroup of order $4$, then the $2$-Sylow subgroup is normal.

So since all the Sylow subgroups are normal, $G$ is isomorphic to the direct product of its Sylow subgroups, $$ G\simeq P_7\times P_2 $$ where $P_7$ is the Sylow $7$-subgroup, and $P_2$ is the $2$-Sylow subgroup. Thus $G$ is abelian.

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