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Center of Factor Group Is Trivial Subgroup

  1. Jul 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that the center of the factor group G/Z(G) is the trivial subgroup ({e}).


    2. Relevant equations
    Z(G) = {elements a in G|ax=xa for all elements x in G}


    3. The attempt at a solution
    I need to prove G is abelian, because G/Z(G) is cyclic, right?
    Then I can say that since G is abelian, Z(G) = G and G/Z(G) is trivial. So the center of the trivial factor group is the trivial subgroup ({e})?

    This is my argument, but I am not sure if it is even headed in the right direction.

    If G/Z(G) is cyclic, then by an element g in G, the generator gZ(G) allows gZ(g)= G/Z(G). Suppose elements a,b in G exist and aZ(G) = (gZ(G))^i = g^i * Z(G)
    and bZ(G) = (gZ(G))^j = g&j * Z(G).

    If a=g^i * x
    and b= g^j * y

    then, ab = (g^i * x) (g^j * y) = g^(i+j) *(xy) = ((g^j)y)((g^i)x) = ba
    Since ab=ba , G is abelian.

    By definition of an abelian group G, and given that Z(G)= {elements a in G|ax=xa for all elements x in G}, we conclude that Z(G) = G. Thereby, if ax=xa, for all elements x in G, then x must be the identity element e. The factor group G/Z(G) contains a single element: the identity element e, which is known as the trivial subgroup. (This last part isn't formalized, how should I go about doing that?)
     
  2. jcsd
  3. Jul 31, 2014 #2

    jbunniii

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    No. If ##G## is abelian, then ##G = Z(G)##, so ##G/Z(G)## itself is trivial (and therefore so is its center). But this is not the general case.

    Writing ##Z## instead of ##Z(G)## for brevity, you need to show that if ##xZyZ = yZxZ## for all ##y \in G##, then ##xZ## is the identity of ##G/Z##, i.e., ##xZ = eZ = Z##, in other words, ##x \in Z##.
     
  4. Jul 31, 2014 #3

    micromass

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    Are you sure this is the correct problem, since it seems false to me.
     
  5. Jul 31, 2014 #4
    I reread it over, and I think it's the same.
    Wait, do you mean it lacks a center, since it only contains e? I thought it was centerless, but I'm not sure.
     
  6. Jul 31, 2014 #5

    jbunniii

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    I was thinking the same thing. Indeed, a counterexample is ##G = D_8##, the dihedral group of order 8. The center of this group has order 2, so ##G/Z(G)## has order 4 and is therefore abelian.
     
  7. Jul 31, 2014 #6
    So, is proving that it is Abelian unnecessary or an incorrect method?
     
  8. Jul 31, 2014 #7

    jbunniii

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    The problem statement is false, so it's not possible to prove it. See my ##D_8## counterexample above.

    The problem statement is (trivially) true if you add the assumption that ##G## is abelian, but that wasn't mentioned anywhere.
     
  9. Jul 31, 2014 #8
    Okay, thank you for the counterexample. But can you not substantiate the claim that G lacks a center?
     
  10. Jul 31, 2014 #9

    jbunniii

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    ##G## does not lack a center. Every group has a center. Indeed ##Z(G)## as given in the problem statement is the center of ##G##.

    ##Z(G)## is the set of elements which commute with every element of ##G##. This is clearly always true of the identity, ##e##, so ##e\in Z(G)## for any group ##G##. So in particular, ##Z(G)## is not empty. It's also straightforward to show that if ##g## and ##h## are elements of ##Z(G)##, then ##gh## and ##g^{-1}## are also elements of ##Z(G)##, which means that ##Z(G)## is a subgroup of ##G##.

    At one extreme, ##Z(G)## might be all of ##G##. This is true precisely when ##G## is abelian. Do you see why?

    At the other extreme, ##Z(G)## might consist only of ##\{e\}##. In this case, we say that ##Z(G)## is trivial. In some sense, ##G## is as NON-abelian as possible in this case, since only the identity commutes with every element of ##G##. An example of a group where ##Z(G)## is trivial is the symmetric group ##S_3##.
     
  11. Jul 31, 2014 #10
    I think I see how Z(G) = G when G is abelian. However, does this not inherently restrict the x in G to only being x= e, or does it change how Z(G) is defined altogether?
     
  12. Jul 31, 2014 #11

    jbunniii

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    No, I'm not sure where you got that idea. Maybe it will be easiest to understand using an example. Can you give an example of a nontrivial abelian group? (Something other than ##G = \{e\}##, which is too trivial to be of much use in explaining anything.)
     
  13. Jul 31, 2014 #12
    S3 is nontrivial and abelian.
     
  14. Jul 31, 2014 #13

    jbunniii

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    No, S3 is not abelian. For example, using cycle notation, (1 2) and (2 3) are both elements of S3, but:

    (1 2)(2 3) = (1 2 3)
    (2 3)(1 2) = (1 3 2)

    Therefore (1 2) and (2 3) do not commute.
     
  15. Jul 31, 2014 #14
    I see now how it is not abelian.
    But, isn't the only way to make the claim in the statement true, to prove that Z(G) is trivial, so that the subgroup is trivial (contains identity element e)?
     
  16. Jul 31, 2014 #15

    jbunniii

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    But the problem statement is not claiming that Z(G) is trivial. It claims that Z(G/Z(G)) is trivial. (Which is false in general.)
     
  17. Jul 31, 2014 #16
    Can you explain how Z(G/(Z(G))) is false?
     
  18. Jul 31, 2014 #17

    jbunniii

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    Do you mean: why is the statement "Z(G/Z(G)) is trivial" is false? Once again, see my counterexample ##D_8##:

    • ##G = D_8##
    • ##Z(G) = \{e, r^2\}## where ##e## is the identity and ##r^2## is the rotation of the square by 180 degrees
    • Therefore, ##Z(G)## consists of 2 elements.
    • By Lagrange's theorem, ##G/Z(G)## must therefore consist of 4 elements
    • Up to isomorphism, there are only two groups of order 4, and they are both abelian: one is ##Z_4##, the cyclic group of order 4, and the other is ##Z_2 \times Z_2##, the direct product of ##Z_2## with itself
    • Therefore, ##G/Z(G)## is abelian
    • Therefore, the center of ##G/Z(G)## is all of ##G/Z(G)##
    • In symbols: ##Z(G/Z(G)) = G/Z(G)##, which has order 4 and therefore is not trivial
     
    Last edited: Jul 31, 2014
  19. Jul 31, 2014 #18
    Sorry, I meant "Why is Z(G/Z(G)) not trivial?"
    But I thought Z(G) by definition has to be abelian since ax=xa for all elements x in G. Is this not a correct conclusion? If x is a variable and x is a different value every time, and a is also different every time, then irregardless ax=xa always because of commutative property of subgroups right?
    Also, I failed to mention that Z(G) is a normal subgroup of G, although I don't think that would change anything.
     
  20. Jul 31, 2014 #19

    jbunniii

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    ##Z(G)## is certainly abelian.
    Well, if ##a## is any element of ##Z(G)## and ##x## is any element of ##G##, then ##ax=xa##. That is true by definition of ##Z(G)##.

    But I'm not sure why you think this should imply that ##Z(G/Z(G))## should be trivial. Can you outline your rationale for why this should be true?

    Yes, this is also true, and it's important because otherwise ##G/Z(G)## would not even be a group, so it would make no sense to talk about its center.
     
  21. Jul 31, 2014 #20
    I think I made a typo when I typed that (I thought G/Z(G) is trivial). I now realize it isn't possible to conclude anything about G/Z(G) unless it is stated that it is cyclic (by which then it is abelian) then the center of the factor group would be trivial since G= Z(G). Thanks for bearing with me and explaining it with such detail.
     
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