0
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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 $K$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.