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# Coset explanation needed please watch

1. Example. If then the subgroup 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 , so it suffices to consider ; then , 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.

Rep awarded on this one
2. 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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