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dpsmith December 3rd, 2017 07:50 AM

Normal Subgroups of p-groups
Let G be a p-group and N ≠ (1) be a normal subgroup of G. Then N and Z(G) have a nontrivial intersection. Is this an induction problem?

johng40 December 3rd, 2017 04:04 PM

Since you mention induction, I assume that G is finite; aside: I realized I don't know if this is true for infinite p-groups. Here's most of a solution that uses a minimal counterexample, a form of induction:

Assume false and let G be a minimal counterexample; i.e. for any p-group of order less than the order of G the statement is true. Now the center of any finite p-group is non-trivial; easy proof via the class equation. So let $Z=Z(G)$ and $Z_2/Z=Z(G/Z)$. By assumption $(NZ\cap Z_2)/Z=(N\cap Z_2)Z/Z$ is not trivial; i.e. $N\cap Z_2$ is non-trivial. Let $1\neq z\in N\cap Z_2$. Then for any $g\in G$, $[g,z]=g^{-1}z^{-1}gzZ=Z$ in $G/Z$. That is $[g,z]\in Z\cap N=<1>$ or $1\neq z\in N\cap Z$, contradiction.

dpsmith December 4th, 2017 02:48 AM

It is assumed to be a finite group - left that out. I thought the Isomorphism THM would be used - but couldn't get the key move.

dpsmith December 15th, 2017 11:33 AM

Found an Easier Way
Actually, you do not need the Isomorphism THM machinery to do this. We have that N is a normal subgroup of G. So if a is in N, every conjugate of a by an element of N is in N.

Hence, constructing a conjugacy class equation for N makes sense. Since N is bigger than (1), you now do the same proof as that which shows that a nontrivial p-group has a nontrivial center. (That is the case when N = G.)

johng40 December 16th, 2017 08:15 AM

Well done. If you want to seriously study groups, factor groups should become second nature for you.

dpsmith December 21st, 2017 05:00 PM

They are. Abstract Algebra was my favorite topic in grad school, which was decades ago. I now am getting into math history and the interesting (and sometimes crazy) people who got involved.

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