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    Example. If G = D_8 then the subgroup H = \{e, a, a^2, a^3\} is a normal subgroup, since its index in G is 8/4 = 2. We may see this directly as follows: clearly ghg^(−1) in H for all g, h \in H , so it suffices to consider g \not\in H; then g = ba^i , h = a^j , and

    ghg^{−1} = ba^i(a^j)(ba^i)^(−1) = (ba^i)(a^j)(a^(−i))(b^(−1)) = b(a^j)(b^(−1)) = (bab^(−1))^j = (a^(−1))^j = a^(−j) in H

    Sorry, I couldn't get the latex code right for the bottom line. However the bit i'm stuck on is the last line:

    b(a^j)(b^(−1)) = (bab^(−1))^j = (a^(−1))^j = a^(−j)

    If anyone can explain how the left progresses to the right hand side, in very obvious terms for me it would be great cause i can't see the woods for the trees atm.

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    b(a^j)(b^(−1)) = (bab^(−1))^j
    just try it:
    eg
    j=1
    [bab^(-1]^1=bab^(-1)

    j=2
    [bab^(-1)]^2=bab^(-1)ba^(-1)b^-1=ba^(-2)b^(-1)

    etc

    (bab^(−1))^j = (a^(−1))^j
    this follows from the group multilplication:
    in D8 you bab^(-1)=a^(-1)
 
 
 
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Updated: February 11, 2010
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