On page 112 of Dixon and Mortimer's Permutation Groups, Theoerm 4.3A (iii) says that every minimal normal subgroup $K$ of $G$ is a direct product $K=T_1 \times \cdots \times T_k$ where $T_i$ are simple normal subgroups of $K$ which are conjugate under $G$.

The proof starts by letting $T$ be a minimal normal subgroup of $K$ and shows that $K$ is a direct product of conjugates of $T$ in $G$.

To show that $T_i$ is simple, it says that "the normal subgroups of $T_i$ are clearly normal in $K$". I can't really see why this is the case...


Because they have already shown that $K$ is the direct product of $T_i$ and some other groups, so $K = T_i \times C$, where $C$ is the direct product of the other $T_j$'s. This means in particular that $C$ centralizes $T_i$. Now suppose that $N$ is normal in $T_i$ and let $k\in K$. Then there are $t\in T_i$ and $c\in C$ such that $k=tc$. Then $$ N^{k} = N^{tc} = N^c = N $$, where the second equality holds because $N$ is normal in $T_i$ and the last equality holds because $c\in C$ centralizes $T_i$ and thus $N\subseteq T_i$. This shows that $N^k = N$ for all $k\in K$, so $N$ is normal in $K$.

  • $\begingroup$ Doesn't $K$ normalises $T_i$ tells us that $k^{-1}T_ik=T_i$? And that's the same as $T_i$ normal in $K$? I still don't see how this leads to normal subgroup of $T_i$ is normal in $K$. $\endgroup$ – BetaY Jun 17 '16 at 7:29
  • $\begingroup$ @JeremyH Yes, $K$ normalises $T_i$ means the same as $T_i$ is normal in $K$. But I now think that this was somewhat misleading. I have rewritten with more details and I hope it is clearer now? $\endgroup$ – ladisch Jun 17 '16 at 8:41
  • $\begingroup$ I think I get it. Thanks! $\endgroup$ – BetaY Jun 17 '16 at 9:28

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