Prove that in a nilpotent group every normal subgroup of prime order is contained in the center.

Question

Prove that in a nilpotent group every normal subgroup of prime order is contained in the center.

Thanks a lot.

score 1 · Accepted Answer · answered Dec 10 '13 at 15:24

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Hints:

Using for example the upper central series

$$1=Z_0\lhd Z_1:=Z(G)\lhd Z_2\lhd\ldots\lhd Z_n=G$$

show that $\;1\neq N\lhd G\implies N\cap Z(G)\neq 1\;$

answered Dec 10 '13 at 15:24

DonAntonio

The result that you told is a standard result. I know it. But with it, how can we prove the main result? – user110834 Dec 10 '13 at 15:26
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Use now that the order of $;N;$ is a prime and the above...that's all! – DonAntonio Dec 10 '13 at 15:27
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Nice work, and great follow up comment, too! +1 – amWhy Dec 10 '13 at 15:29
Thanks @amWhy. I dealt a lot with groups, nilpotent and all, while in graduate school: now it pays! :) – DonAntonio Dec 10 '13 at 15:30
Hmmm...I have yet to be "paid" for my work here at MSE! ;-) – amWhy Dec 10 '13 at 15:31
We have the order of $N$ is a prime, and $N\cap Z(G)\neq 1$. But why $N$ is contained in $Z(G)$? – user110834 Dec 10 '13 at 15:31
Can you explain more precisely? – user110834 Dec 10 '13 at 15:33
Hmmm...take $;1\neq x\in N\cap Z(G);$, @user110834: what is $;\langle x\rangle;$ ?? – DonAntonio Dec 10 '13 at 15:33
I need more details. May you give me? – user110834 Dec 10 '13 at 15:40
@user110834, what's the order of any non trivial element in a group of prime order?!? – DonAntonio Dec 10 '13 at 15:42
Ehm, of course. I forgot the order of $N$ is a prime. :) Thank you very much. – user110834 Dec 10 '13 at 15:43

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