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I would appreciate a hint on how to go about solving the following problem.

Let $X$ be a finite set of minimal normal subgroups of $G$ and write $K:=\prod X$ (the product of all members of $X$).

a) Show that $K$ is the direct product of some subcollection $Y \subseteq X$.

b) If $Y$ is as in (a), show that all members of $X \setminus Y$ are contained in $Z(K)$. Conclude that $Y$ contains all nonabelian members of $X$.

Let $X=\{N_1,...,N_k\}$. I have the result that each $N_i$ is the direct product of isomorphic (specifically, conjugate in $G$) simple subgroups. Since the center of a direct product is the direct product of the centers, this tells us that for all $N_i$, either $Z(N_i)=1$ or $Z(N_i)=N_i$. This plus part (b) seems to indicate that we can leave out the abelian subgroups, but I certainly do not see why.

If the $N_i$ are finite, then we must have $Y=X$ because otherwise $\prod Y$ must have smaller order and cannot equal $K$, so I think I should be working in the fully general case of infinite minimal normal subgroups in infinite groups.

Is there another property of minimal normal subgroups that I'm not thinking of but should be considering?

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a) Take any $X_1$ in $ X$. If $X_1=K$, it is done. If not, there exists $X_2$ in $X$ such that $X_2\not \le X_1$. Now consider $N=X_1\cap X_2$. Clearly $N \lhd G$. By the minimality of $X_i$, we get $N=1$. Hence $X_1X_2=X_1 \times X_2$. In this way, we can get $K$ is direct product.

b) Now let $Y=\{X_1, \cdots, X_r\}$. Take any $Y_i \in X -Y$. Then $X_i \ne Y_j$. Clearly, $[X_i, Y_j]$ is normal in $G$ contained in $X_i $ and $Y_j$. By the minimality of $X_i, Y_j$, we see that $[X_i, Y_j]=1$, which means $Y_j \le Z(K)$.

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  • $\begingroup$ For (a): I see that $K$ is a direct product of the $X_i$, but how can it be written as the direct product of some subcollection of $X$? $\endgroup$ – Alex Petzke Feb 21 '14 at 3:12
  • $\begingroup$ These $X_i$ are choose from X. $\endgroup$ – Wei Zhou Feb 21 '14 at 4:00
  • $\begingroup$ Yes, I understand that. But when will it be the case that $Y$ is a proper subcollection in a 'nontrivial' way, such as $X$ containing multiple copies of the same subgroup? I just don't see the significance of that construction. $\endgroup$ – Alex Petzke Feb 25 '14 at 2:07
  • $\begingroup$ For example, G is the elementary abelian group of order $4$, generated by a and b. Then X={<a>,<b>,<ab>}, Y={<a>,<b>}. This is a nontrivial example, if I understand your comment correctly. $\endgroup$ – Wei Zhou Feb 25 '14 at 5:38
  • $\begingroup$ I see. Thanks for the help. $\endgroup$ – Alex Petzke Feb 26 '14 at 2:43

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