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Herstein, Topics in Algebra, page 58

  1. Mar 15, 2008 #1
    I have the second printing of the first edition of Herstein's 'Topics in Algebra', published 1964.

    On page 58 near the middle of the page there is a paragraph that begins:

    Let G be a cyclic group ...

    The author writes
    [tex]\phi:a^i \rightarrow a^{2i}[/tex]

    and later

    [tex]x^{-1}a^ix = \phi(a)^i = a^{3i}[/tex]

    The next paragraph makes it clear that he means:
    [tex]x^{-1}a^ix = \phi^i(a) = a^{3i}[/tex]

    But it doesn't seem true to me. for instance if i = 1, then no matter how I write it, I get:
    [tex]\phi(a) = a^3[/tex]

    but by the definition of phi,
    [tex]\phi(a) = a^2[/tex]

    What gives?
     
  2. jcsd
  3. Mar 15, 2008 #2

    HallsofIvy

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    What is x in [tex]x^{-1}a^ix = \phi(a)^i = a^{3i}[/tex]? Surely it can't be just any member of G because then we would have a= a3.

    And what is a? Any member of G or specifically a generator of G?
     
  4. Mar 15, 2008 #3
    Sorry, I didn't put enough information for anyone that doesn't have a copy of the book. G is a cyclic group of order 7, a is an element of G, so that [itex]G = \{e = a^0, a^1, a^2, a^3, a^4, a^5, a^6\}[/itex]. x is a formal symbol. The author intends to describe the group of order 21 made of formal symbols [itex]x^ia^j, i = 0, 1, 2 j = 0, 1, 2, 3, 4, 5, 6[/itex].
     
  5. Mar 16, 2008 #4

    morphism

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    I think it's a typo and should be [itex]a^{2i}[/itex] instead. Unfortunately I don't have my copy of Herstein on me right now to verify this. Maybe you could post a bit more of that page?

    His intent is clear though: he's trying to define the semidirect product of G and X={1, x, x^2}, with X viewed as the cyclic group of order 3, where conjugation by x acts as [itex]\phi[/itex] on G.
     
  6. Mar 17, 2008 #5
    Perhaps. However, he gives a specific example of multiplication in the larger group.
    [tex]x^1a^1 \cdot x^1a^2 = x^1(a^1x^1)a^2 = x^1(x^1a^3)a^2 = x^2a^5[/tex]
    That's taking a typo pretty far, but I suppose it's possible he lost track half way through the paragraph.
     
  7. Aug 30, 2008 #6
    It was a typo in the 1st edition.
    The corrected expression is on pg 69 of the 2nd edition:

    x^{-1}a^ix = \\phi(a^i) = a^{2i}
     
  8. Aug 30, 2008 #7
    With formatting....
    [tex]
    x^{-1}a^ix = \phi(a^i) = a^{2i}
    [tex]
     
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