About minimal normal groups and subnormal groups

Hello everybody, I would like to have some help with these two questions:

1. Let $\displaystyle G$ be a finite nilpotent group and $\displaystyle N$ a minimal normal subgroup. Show that $\displaystyle N\leq Z(G)$.

2. Let $\displaystyle G$ be a group and $\displaystyle S,T$ verifying $\displaystyle S\neq T$ non-abelian subnormal subgroups of $\displaystyle G$. Prove that $\displaystyle st=ts$ $\displaystyle \forall s\in S$ $\displaystyle \forall t\in T$

1. I have already proved that $\displaystyle N$ must be abelian; $\displaystyle N\leq Z(N)$, but I find no way of proving that $\displaystyle N\leq Z(G)$ using that the nilpotency of $\displaystyle G$ and each one of its characterizations. Any idea?

Regards and thanks in advance.

Sheila.

Re: About minimal normal groups and subnormal groups

For the first one, a group G is nilpotent is also solvable. So, the derived series

$\displaystyle G^{(1)}, G^{(2)},\dots$

terminates for some $\displaystyle k$.

I suggest using the correspondence theorem for factor groups and the fact that $\displaystyle G^{(i)}/G^{(i+1)}$ is abelian. There will be something you need to show about $\displaystyle N$ though.

Re: About minimal normal groups and subnormal groups

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Originally Posted by

**Sheila496** 1. Let $\displaystyle G$ be a finite nilpotent group and $\displaystyle N$ a minimal normal subgroup. Show that $\displaystyle N\leq Z(G)$.

$\displaystyle N, Z(G)\unlhd G \implies N \cap Z(G) \unlhd G\,\, \overset {N \cap Z(G)\leq N}{\implies}\,\, N \cap Z(G) \unlhd N$.

But$\displaystyle N$ is a minimal normal subgroup so $\displaystyle N \cap Z(G)= N \implies N\leq Z(G)$.

Re: About minimal normal groups and subnormal groups

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**zoek** $\displaystyle N, Z(G)\unlhd G \implies N \cap Z(G) \unlhd G\,\, \overset {N \cap Z(G)\leq N}{\implies}\,\, N \cap Z(G) \unlhd N$.

But$\displaystyle N$ is a minimal normal subgroup so $\displaystyle N \cap Z(G)= N \implies N\leq Z(G)$.

But you have another possibility: $\displaystyle N\cap Z(G)=1$. This is the difficult case, where you have to find a contradiction.

Thanks anyway.

Re: About minimal normal groups and subnormal groups

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Originally Posted by

**Haven** For the first one, a group G is nilpotent if its solvable. So, the derived series

$\displaystyle G^{(1)}, G^{(2)},\dots$

terminates for some $\displaystyle k$.

I suggest using the correspondence theorem for factor groups and the fact that $\displaystyle G^{(i)}/G^{(i+1)}$ is abelian. There will be something you need to show about $\displaystyle N$ though.

Thanks, that may help me finding a proof. But it seems like you want to prove it using that $\displaystyle G$ is solvable, however in this case the result is false: $\displaystyle \Sigma_4$ is solvable and $\displaystyle 1<V_4\cdot\unlhd \Sigma_4$, but $\displaystyle Z(\Sigma_4)=1$. That shows that you have to use somewhere the nilpotency of $\displaystyle G$

Re: About minimal normal groups and subnormal groups

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Originally Posted by

**Sheila496** But you have another possibility: $\displaystyle N\cap Z(G)=1$. This is the difficult case, where you have to find a contradiction.

Thanks anyway.

But, this can't happen. The center of nilpotent groups are "normality-large" in the sense that they intersect non-trivially with every normal subgroup. This can be found in any good text on group theory (e.g. Isaac's or Robinson).

Re: About minimal normal groups and subnormal groups

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**Drexel28** But, this can't happen. The center of nilpotent groups are "normality-large" in the sense that they intersect non-trivially with every normal subgroup. This can be found in any good text on group theory (e.g. Isaac's or Robinson).

Thank you. Certainly that was my main purpose. It's known that the behaviour of nilpotent groups is "similar" to $\displaystyle p$-groups, and that happens in these last ones. I'll look for a proof, but it seems they all use commutators, this will take some time for me because I have never worked with them.

Re: About minimal normal groups and subnormal groups

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Originally Posted by

**Sheila496** Thank you. Certainly that was my main purpose. It's known that the behaviour of nilpotent groups is "similar" to $\displaystyle p$-groups, and that happens in these last ones. I'll look for a proof, but it seems they all use commutators, this will take some time for me because I have never worked with them.

What do you mean commutators, the fact that $\displaystyle [G,\mathcal{Z}_{n+1}(G)]\leqslant\mathcal{Z}_n(G)$? A proof can be found here--that's a good website, bookmark it.

Re: About minimal normal groups and subnormal groups

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**Drexel28** What do you mean commutators, the fact that $\displaystyle [G,\mathcal{Z}_{n+1}(G)]\leqslant\mathcal{Z}_n(G)$? A proof can be found

here--that's a good website, bookmark it.

Yes, I found that webpage before posting the problem here, but I didn't know the meaning of $\displaystyle [H,K]$ being $\displaystyle H,K\leq G$. Now I'll take a look at the books you recommended in order to understand completely the proof given there. Interesting work indeed.

Re: About minimal normal groups and subnormal groups

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Originally Posted by

**Sheila496** Yes, I found that webpage before posting the problem here, but I didn't know the meaning of $\displaystyle [H,K]$ being $\displaystyle H,K\leq G$. Now I'll take a look at the books you recommended in order to understand completely the proof given there. Interesting work indeed.

Good luck friend, but just for soundness the map $\displaystyle [\cdot,\cdot]:G\times G\to G$ is given by $\displaystyle [g,h]=ghg^{-1}h^{-1}$. Then one defines $\displaystyle [H,K]$ to be the image of $\displaystyle H\times K$ under $\displaystyle [\cdot,\cdot]$.