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Need help! Inverse!

  1. Sep 15, 2008 #1
    How would one prove that if h is in H, then also h^-1, that is its inverse of it is also in H, if Z is only positive integers.
    Where [tex]H=\{a^n: n\in Z^+\}[/tex] a in G
    I managed to show, as stated that [tex] e=a^{m-n}\in H[/tex]
    since i supposed that the group G is finite, and i also know that now i have to take two elements in H, say h and h' and then use the property that H is closed, and [tex]hh'=...=e\in H[/tex]

    , but i cannot figure out how to pick up h and h'?

    Can u help me on this?
    Last edited: Sep 15, 2008
  2. jcsd
  3. Sep 16, 2008 #2


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    Hi sutupidmath! :smile:

    (I'm not sure I understand what G and H are, but in case I've got it right:)

    Hint: if G is finite, and Hh = {hn : n ε Z+}, then how large is Hh? :wink:
  4. Sep 16, 2008 #3
    I think i got it, let's see:

    SInce [tex] g^{m-n}=e\in H[/tex] it means that H consists of m-n elements, that is all powers of g up to m-n. Now this means that there defenitely are:

    [tex] s_1, s_2 \in Z^{+}[/tex] such that [tex] s_1+s_2 =m-n[/tex] while [/tex] g^{s_1},g^{s_2} \in H[/tex] this means that:

    For every element in H, say [tex] h=g^{s_2}, s_2 \in Z^+, \exists h'=g^{s_1}, s_1 \in Z^+[/tex] such that

    using closure property of H, we get:

    [tex] hh'=g^{s_1}g^{s_2}=g^{s_1+s_2}=g^{m-n}=e \in H[/tex] which means that h and h' are inverses of each other, and still both h and h' are in H. Proof done. RIght?
  5. Sep 17, 2008 #4


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    Hi sutupidmath! :smile:

    This is very confusing …

    I take it you're using g instead of a? …

    and I don't understand what m is, or why you're using m - n when n is just an index. :confused:

    Can't you shorten it, by starting with something like "for any s2 ε Z+, define s1 = -s2 mod m - n …" ? :smile:
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