Let $G$ be a finite group and $(*)$ be the property:

$(*)$: Every minimal normal subgroup is contained in the center.

$(a)$ Let $N$ and $M$ be normal subgroups of $G$, both of which satisfy $(*)$. Then prove: $NM$ satisfies $(*)$.

$(b)$ If $G$ satisfies $(*)$, then prove: every normal subgroup of $G$ satisfies $(*)$.

I would also agree that it may be a possible duplicate of this post. However, unfortunately, I could find that post a little complicated to understand and not well developed, of which the logic, notation, and language might not have been so polished.

It would be greatly appreciated, if you could throw light on this question and be kind enough to give an elegant proof. Thanks a lot! $\ddot\smile$


I am not sure that I can achieve elegance.

Let's deal with (b); then (a) is a corollary. So we have a normal subgroup $N$ of $G$.

Step 1: Let $K$ be a minimal normal subgroup of $N$ with $K\not\leqslant Z(N)$. Then $K\cap Z(N)=1$ or we are done.

Step 2: Consider $K^G$. It is not hard to prove that this is a normal subgroup of $G$, and indeed that $K^G=K_1\times K_2\times\dots\times K_s$, where $K_1=K$ and each $K_i$ is a $G$-conjugate of $K$, and each is normal in $N$.

Step 3: Let $L$ be a minimal normal subgroup of $G$ lying in $K^G$. Then by hypothesis $L\leqslant Z(G)$, and so $L\leqslant Z(N)$.

Step 4: Let $1\not=(x_1,x_2,\dots,x_s)\in L\cap Z(N)$, where $x_i\in K_i$. Now $N$ fixes each $K_i$, so we have that $N$ centralises each $x_i$.

Step 5: Some $x_i\not=1$, so we get $K_i\cap Z(N)\not=1$. Conjugating by an appropriate element of $G$ we get that $K_1\cap Z(N)\not=1$; this is a contradiction, so $K\leqslant Z(N)$.

[I think Step 2 is standard stuff.]

  • $\begingroup$ Why could we claim $L$ to be in $K^G$? $\endgroup$ – user542899 Mar 22 '18 at 8:27
  • $\begingroup$ $K^G$ is a normal subgroup of $G$ so we can choose a minimal normal subgroup inside it. $\endgroup$ – ancientmathematician Mar 22 '18 at 8:45
  • $\begingroup$ Ha! Elegant indeed! $\endgroup$ – Math Mar 22 '18 at 8:51
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    $\begingroup$ Step 3: Don’t we generally have $K^G=N$? (since $K^G\trianglelefteq G, K^G\subseteq N$ and $N$ being minimal normal in $G$) $\endgroup$ – Math Mar 23 '18 at 1:52
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    $\begingroup$ @Benny: $N$ is any old normal subgroup. $\endgroup$ – ancientmathematician Mar 23 '18 at 7:39

I have an argument for (a) in the case $N \cap M = 1$, which implies $NM = N \times M$. Let $K$ be some minimal normal subgroup of $NM$. Then by considering the quotient $KN / N \unlhd NM / N \cong M$ either $KN \le N$ or $KN / N$ is minimal normal in $MN / N$, in the last case $[x,y] \in N$ for each $x \in K, y \in NM$ by assumption. Similar by looking at $NM / M$ we find $K \le M$ or $[x,y] \in M$ for each $x \in K, y \in NM$. As $[N, G] \le N$ and $[M, G] \le M$ by normality, in all cases we have $[x,y] \le M \cap N = 1$ for $x \in K, y \in NM$, which gives the claim.

If $N \cap M \ne 1$ maybe (b) can help in the sense that we can deduce that property $(*)$ then holds for $M \cap N$.

EDIT (5/11/19): I found a proof for the general case which works quite different. Let $K$ be some minimal normal subgroup in $NM$. Then as $N$ is normal we have that $[K, N] \le N \cap K$ is normal in $NM$, which implies $[K,N] = 1$ or $[K,N] = K$. Assume $[K,N] = K$. Let $L \le K$ be some minimal normal subgroup of $N$ in $[K,N]$ and consider $H = \langle L^g \mid g \in NM \rangle$, which is normal in $NM$. Hence $H = [K,N] = K$. The groups $L^g$ are also minimal normal in $N$ by normality of $N$ and the fact that non-trivial homomorphic images are minimal normal in its image, hence $H$ is a direct product of some subset of these conjugages. By assumption the factors are all in the center of $N$, hence $K$ is in the center of $N$, i.e. $[K,N] = 1$ contradicting the assumption. Simlar we find $[K,M] = 1$. So $[K,MN] = [K,M][K,N] = 1$, i.e. $K$ is in the center of $NM$. $\square$


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