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Thread: Nilpotent Groups w/ Normal Subgroups

  1. #1
    Senior Member roninpro's Avatar
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    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.
    Last edited by roninpro; Nov 28th 2009 at 10:11 PM. Reason: Incorrect tags for LaTeX formatting. Corrected the statement of the problem.
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  2. #2
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    Quote Originally Posted by roninpro View Post
    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 $\displaystyle g \in N \cap Z(G). \ \Box$
    Last edited by NonCommAlg; Nov 28th 2009 at 09:30 PM.
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  3. #3
    Senior Member roninpro's Avatar
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    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.
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