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My book says: Since $B$ is normal, the quotient group $AB/B$ is well defined. I want to know why the "well-definedness" of $AB/B$ is dependent on $B$ being normal. Let me know if you need more context to make sense of this question. Also, why is the second isomorphism theorem also called the diamond isomorphism theorem? Is it a historical thing?

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One always needs a subgroup to be normal in order for the set of its cosets to form a group in the natural way. The point is that being normal in the ambient group $G$ implies being normal in the subgroup $AB.$ –  Andrew Dec 11 '12 at 18:36
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up vote 2 down vote accepted

The quotient group is defined to be the set of all (let's say left) cosets of $B$, written $gB$ for some $g \in G$ with the opertation defined as $(gB) \circ (hB) = (g\circ h)B$. For this to be well defined, we require that $gB = g'B$ and $hB = h'B$ $\implies ghB = g'h'B$. That is to say, any two representations of the same coset behave in the same way, so it doesn't matter which one we choose.

This is where normality comes in, $gB=g'B \implies\exists k_1 \in B$ such that $g' = gk_1$. Similarly $\exists k_2$ such that $h' = hk_2$. Now, $$g'h'B = gk_1hk_2B = g(hh^{-1})k_1hk_2B$$ normality of $B$ means that $h^{-1}k_1h = k_3$ for some $k_3 \in B$. Hence from above:

$$g'h'B = ghk_3k_2B = ghB$$

This shows that if $B$ is normal, then the quotient group is well defined. It's also possible to prove that if the quotient group is well-defined then $B$ is normal. This isn't too hard, and I hope that the above is more what you were after anyway. If not let me know and I'll fill in the rest of the details.

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Nope, that was very useful. Thanks! –  chubbycantorset Dec 11 '12 at 18:58
    
@chubbycantorset no problem, happy to help! –  Tom Oldfield Dec 11 '12 at 19:00
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