# Thread: Nilpotent Groups w/ Normal Subgroups

1. ## Nilpotent Groups w/ Normal Subgroups

Hello. I have been asked to prove the following:

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

I attempted to take the factor group $G/N$ (since it too is nilpotent) and try to deal with the image of $Z(G)$ 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 $N$ is a normal subgroup of a nilpotent group $G$, then $N \cap Z(G)$ is nontrivial.

I attempted to take the factor group $G/N$ (since it too is solvable) and try to deal with the image of $Z(G)$ 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 $N \neq \{1 \}.$ let $\{1 \}=G_0 \subset G_1 \subset \cdots \subset G_n=G$ be the upper central series of $G.$ since $N \cap G_n=N \neq \{ 1 \},$ we can define $k=\min \{j : \ N \cap G_j \neq \{ 1 \} \}.$

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

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

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.