# Thread: Nilpotent Groups w/ Normal Subgroups

1. ## Nilpotent Groups w/ Normal Subgroups

Hello. I have been asked to prove the following:

If $\displaystyle N$ is a nontrivial normal subgroup of a nilpotent group $\displaystyle G$, then $\displaystyle N \cap Z(G)$ is nontrivial.

I attempted to take the factor group $\displaystyle G/N$ (since it too is nilpotent) and try to deal with the image of $\displaystyle Z(G)$ under the canonical homomorphism, but I can't get the result to come out.

This is extremely frustrating, and thus, I would appreciate any suggestions you may have. (Though, I am not necessarily searching for a complete solution.)

Thank you.

2. Originally Posted by roninpro
Hello. I have been asked to prove the following:

If $\displaystyle N$ is a normal subgroup of a nilpotent group $\displaystyle G$, then $\displaystyle N \cap Z(G)$ is nontrivial.

I attempted to take the factor group $\displaystyle G/N$ (since it too is solvable) and try to deal with the image of $\displaystyle Z(G)$ under the canonical homomorphism, but I can't get the result to come out.

This is extremely frustrating, and thus, I would appreciate any suggestions you may have. (Though, I am not necessarily searching for a complete solution.)

Thank you.
you forgot to mention the condition $\displaystyle N \neq \{1 \}.$ let $\displaystyle \{1 \}=G_0 \subset G_1 \subset \cdots \subset G_n=G$ be the upper central series of $\displaystyle G.$ since $\displaystyle N \cap G_n=N \neq \{ 1 \},$ we can define $\displaystyle k=\min \{j : \ N \cap G_j \neq \{ 1 \} \}.$

note that $\displaystyle k \geq 1$ because $\displaystyle N \cap G_0=\{ 1 \}.$ so $\displaystyle N \cap G_k \neq \{1\}$ and $\displaystyle N \cap G_{k-1} = \{ 1 \}.$ choose $\displaystyle 1 \neq g \in N \cap G_k.$ then, since $\displaystyle gG_{k-1} \in G_k/G_{k-1} = Z(G/G_{k-1}),$ we have $\displaystyle gxG_{k-1}=xgG_{k-1},$ for all

$\displaystyle x \in G.$ therefore $\displaystyle y=g^{-1}x^{-1}gx \in G_{k-1}.$ but, since $\displaystyle N$ is normal and $\displaystyle g \in N,$ we also have $\displaystyle y \in N.$ thus $\displaystyle y \in N \cap G_{k-1}=\{1\}.$ that means $\displaystyle gx=xg,$ for all $\displaystyle x \in G.$ hence $\displaystyle g \in N \cap Z(G). \ \Box$

3. Sorry for the mistakes - I typed up the post in a hurry.

I originally went about halfway in the direction of your solution but couldn't see what to do with it, so I discarded it, unfortunately.

I see how to do it now. Thank you very much for your response.