Theorem: Let N be a normal subgroup of G, and let G/N denote the set of all righ

Question

Theorem: Let N be a normal subgroup of G, and let G/N denote the set of all right
cosets of N in G. For Na in G/N and Nb in G/N, let (Na)(Nb) = N(ab).
With this operation G/N is a group called the quotient group (or factor group) of G by N.

The operation on G/N is associative because if a, b, c in G, then
Na(NbNc) = Na(Nbc) = N(a(bc)) = N((ab)c) = N(ab)Nc = (NaNb)Nc.
The element Ne is an identity element because if a ? G, then
NeNa = N(ea) = Na and NaNe = Nae = Na.
Finally, Na^-11 is an inverse for Na because
NaNa^-1 = N(aa^-1) = Ne and Na?1Na = N(a^-1a) = Ne.
This proves that G/N is a group.

Give a reason for each step in the proof of the theorem for why the operation on G/N is associative

Explanation / Answer

proof of the therom with every step explained : part 1 = http://lookpic.com/O/i2/1509/HfQddPwE.png part 2 = http://lookpic.com/O/i2/1114/W6m0A21J.png

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