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I'm having some problems with understanding the corollary of the next theorem :

Every minimal normal subgroup $K$ of $G$ is a direct product $K = T_1 \times T_2 \times \cdots \times T_k$ where the $T_i$ are simple normal subgroups of $K$ which are conjugate under $G$.

The corollary states :

Every minimal normal subgroup of a finite group is either an elementary abelian p-group for some prime p, or its centre is equal to 1.

I don't understand what from the theorem implies the statement of the corollary. If someone could give me a hint I would be thankful.

Thanks!

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Does the "its" in the corollary refer to the finite group, or to the minimal normal subgroup of the group? –  Arturo Magidin Jan 10 '12 at 15:06

1 Answer 1

up vote 2 down vote accepted

If $K$ is a minimal normal subgroup of $G$, then it is a direct product of isomorphic copies of a simple group (they have to be isomorphic, because they are conjugate in $G$).

One possibility is for this simple group to be cyclic of prime order; in that case, $K$ is a direct product of finitely many cyclic groups of order $p$, and this is called an "elementary abelian $p$-group".

What if the simple group is not abelian? Then $K = S\times S\times\cdots\times S$, where $S$ is a nonabelian simple group. The center of a direct product is the direct product of the centers, and the center of a nonabelian simple group is trivial (since the center is normal, $Z(S)=\{1\}$ or $Z(S)=S$, and since $S$ is nonabelian, $Z(S)\neq S$). Thus, the center of $K$ is trivial.

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Thanks, I got it now. –  aldo Jan 10 '12 at 15:15

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