{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw4_solutions - Solutions to Homework 4 May 1 2010 1 We...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Homework 4 May 1, 2010 1. We will prove this via induction: The first term in the Upper Central Series is { e } , which is clearly a characteristic group. Suppose, for proof by mathematical induction, we have shown that { e } = Z 0 Z 1 ≤ · · · ≤ Z k are all characteristic subgroups of G . Let us now consider Z k +1 : Let ψ : G G/Z k be the obvious mapping (note that Z k is normal in G ). Notice that since Z k is characteristic, we have that σ will map the kernel of ψ into itself; in other words, ψ ( x ) = e ⇐⇒ ( ψσ )( x ) = e. (1) Let Z = Z ( G/Z k ) Then, Z is normal in G/Z k , and its inverse image via ψ is normal in G , and is what we call Z k +1 . Now suppose z Z k +1 . We will now attempt to show that ( ψσ )( z ) commutes with everything in G/Z k , which would prove ( ψσ )( z ) Z , and therefore σ ( z ) Z k +1 ; applying the same argument using σ 1 , which is also an automorphism of G , we will get σ ( Z k +1 ) = Z k +1 , thereby completing the proof of the induction step.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern