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Let $N$ be a non-Abelian minimal normal subgroup of the finite group $G$, and let $\mathcal{K}$ be the set of minimal normal subgroups of $N$, then show that the elements of $\mathcal{K}$ are non-Abelian simple groups, which are conjugate in $G$.

I managed to show that all elements in $\mathcal{K}$ are non-Abelian simple groups, but I was stuck when it comes to showing that they are conjugate.

That they are conjugate means that the factors of the direct product would be uniquely determined, but how to deal with it?

$\ddot\smile$ Any help will be sincerely appreciated! Thanks!

PS: It’s 1.7.5 $(a)$ on page 38 of my textbook, The Theory of Finite Groups, An Introduction. It says that we can get the uniqueness from 1.6.3 $(b)$. But I find that it only helps me with 1.7.5 $(b)$. So what have I missed?

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It follows from the result in your previous question that $N = E_1 \times \cdots \times E_k$, where the $E_i$ are nonabelian simple groups that are all conjugate in $G$.

So all that remains to prove is that the $E_i$ are the only minimal normal subgroup of $N$. To do that, suppose that $M$ is some minimal normal subgroup, and let $1 \ne h \in M$. Then we can write $h$ (uniquely) as $h_1h_2\ldots h_k$ with $h_i \in E_i$. Since $1 \ne h$, some $h_i \ne 1$. Since $E_i$ is nonabelian simple, $Z(E_i)=1$, so there exists $g \in E_i$ with $g \not\in C(h)$. The $1 \ne [g,h] \in M \cap E_i$, and since $E_i$ is simple and $M \unlhd G$, we have $E_i \le M$ and then minimality of $M$ gives $M = E_i$.

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    $\begingroup$ It’s so kind of you to give nice answers all the time! $\ddot\smile$ $\endgroup$ – Benny Mar 15 '18 at 16:17

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