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Suppose that $A$ and $B$ are normal subgroups of $G$ such that they are minimal among all non-trivial normal subgroups of $G$ (minimal with respect to the $\subseteq$ order). Prove that $AB$ is Abelian.

Here's my chain of thoughts so far:

Since both $A$ and $B$ are normal in $G$, their intersection $A \cap B$ is normal in $G$ too. By a minimality argument, we conclude that two cases are possible:

  1. $A \cap B = \{e\}$
  2. $A = A \cap B = B$

We know that:

$$ \frac{AB}{A} \cong \frac{B}{A \cap B}$$ $$ \frac{AB}{B} \cong \frac{A}{A \cap B}$$

If $A = A \cap B =B$, then $\displaystyle \frac{AB}{A} \cong \frac{AB}{B} \cong \{e\}$. But this means that $A=AB=B$. Now I'm stuck. :/

If $A \cap B = \{e\}$, then $\displaystyle \frac{AB}{A} \cong B$ and $\displaystyle \frac{AB}{B} \cong A$. Again, I'm stuck. \:

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    $\begingroup$ Not true. Take $A = A_5 \times 1$, $B = 1 \times A_5$ in $G = A_5 \times A_5$. This would be true if you assumed that $G$ is solvable. $\endgroup$
    – spin
    May 11 '14 at 8:14
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As spin says, I am writing the solution by assuming $G$ is solvable.

claim1: Let $M$ be minimal normal subgroup of $G$, then $M$ is abelian.

Since $G$ is solvable then $M$ is solvable so $M'$ is a proper subgroup of $M$. But notice that $M'$ is a characteristic group in $M$ and $M$ is normal in $G\implies M'$ is normal in $G$. By minimality of $M$, $M'=1 \implies M$ is abelian.

Now let $A,B$ be minimal normal subgroups of $G$ then we know them both of them is abelian.

as you wrote, if $A\cap B=1 \implies AB\cong A\times B$ which must be abelian.

if $A\cap B=B \implies A\subseteq B \implies AB=B$ which is abelian.

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  • $\begingroup$ Why $M'$ is a proper subgroup of $M$? Is it by definition of solvability or something? $\endgroup$
    – math.n00b
    May 11 '14 at 11:19
  • $\begingroup$ @math.n00b: Notice that if $M=M'$ then $M=M''$ then $M=M'''$... which means that $M$ is not solvable. In solvable groups, $M'$ must be proper in $M$. $\endgroup$
    – mesel
    May 11 '14 at 11:27
  • $\begingroup$ Ah, Right! Thanks. $\endgroup$
    – math.n00b
    May 11 '14 at 11:31
  • $\begingroup$ @you are welcome. $\endgroup$
    – mesel
    May 11 '14 at 11:33

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