1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is my approach to this problem correct?

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose H is a normal subgroup G and H has order 2. Show that H is contained in the center of G.

    2. Relevant equations

    Definitions.

    H is a normal subgroup of G if for all h in H and g in G, ghg-1 is in H.

    The center of a group G is Z = {z in G | zx = xz for all x in G}.

    3. The attempt at a solution

    |H| = 2 implies H = {1, h = h-1} for some h. If, in addition, H is a normal subgroup, then for any g in G we have ghg-1 in H = {1, h}. I need to deduce from this that gx = xg for all x in G. Am I right? Is there something missing?
     
  2. jcsd
  3. Oct 8, 2012 #2

    Zondrina

    User Avatar
    Homework Helper

    By "H is a normal subgroup G", do you mean H≤G or H<G? A bit of notational differences here.
     
    Last edited: Oct 8, 2012
  4. Oct 8, 2012 #3
    The definition I posted is straight from the book. But yea, it would mean H is a subset of G.
     
  5. Oct 8, 2012 #4

    Zondrina

    User Avatar
    Homework Helper

    Ahh I just wanted to make sure you didnt mean a proper subset.

    Instead of 1, you should argue a general identity element e in H. Since H is a subgroup of G, you know that H has an identity element e you can use. Which also has an inverse :).

    So, use these to aid you in showing that [itex]e \in H[/itex] and H a subgroup of G, imply that e is also in the center of G. Since the identity element commutes with every other element right?
     
  6. Oct 8, 2012 #5

    I like Serena

    User Avatar
    Homework Helper

    Hi Jamin2112! :smile:

    Seems to me it's a bit sharper.
    You have only 1 h in H, so for that h, you should have that ##\forall g: gh=hg##.
    And since H is normal in G.....
     
  7. Oct 8, 2012 #6
    So inevitably H = {e, e-1}?
     
  8. Oct 8, 2012 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    e^(-1)=e, yes? H has to contain TWO elements. They can't BOTH be e.
     
  9. Oct 9, 2012 #8
    Right. e is its own inverse, so to speak.
     
  10. Oct 9, 2012 #9
    Ok. Here's where I am.

    Suppose H has order 2. By the definition of a group, clearly H = {e, h} where e is the identity element in H and h is some other element. Now suppose that H is a normal subgroup of G. Choose g in G. We have ghg-1 and geg-1 both in H. These are distinct elements, the distinct elements in H commute. Hence ghg-1(geg-1) =geg-1(ghg-1).

    How wrong am I?
     
  11. Oct 9, 2012 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You aren't really thinking this through before posting yet another guess at the answer and waiting for someone to correct you. I think that's a pretty bad approach. What is geg^(-1)? Really this time, no guessing allowed.
     
  12. Oct 9, 2012 #11
    geg^(-1) = e
     
  13. Oct 9, 2012 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    And so? What about ghg^(-1)? Could that be e??
     
  14. Oct 9, 2012 #13
    No. It would have to be h.
     
  15. Oct 9, 2012 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You should probably give a reason for that. And once you know that, how does it help you solve your problem?
     
    Last edited: Oct 9, 2012
  16. Oct 9, 2012 #15
    You right multiply by g to obtain gh = hg.
     
  17. Oct 9, 2012 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, you can write up a proof now, yes?
     
  18. Oct 9, 2012 #17
    I just wanna be sure of one more thing: How do I know e is G's identity element? I know it's H's identity element and I know it is contained in G, but that doesn't necessarily mean it's G's identity element, no?
     
  19. Oct 9, 2012 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Odd question. If h is any element of H then hh^(-1)=e, where e is the identity of G. So e MUST be in H. Look at your definition of 'subgroup'.
     
    Last edited: Oct 9, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is my approach to this problem correct?
Loading...