What does the quotient group $(A+B)/B$ actually mean?

Question

I understand that $A+B$ is the set containing all elements of the form $a+b$, wit $a\in A, b\in B$.

When you do the quotient group, that's like forming equivalence classes modulo $B$. All elements in $B$ should be congruent to 0 modulo $B$, as far as I understand. So wouldn't $(A+B)/B$ be the same as $A/B$?

Where is the error in my reasoning?

Answer 1

It is the same if $B\subset A$, while if $B\not\subset A$, $A/B$ isn't even defined. Actually $A+B$ is the smallest subgroup that contains both $B$ and $A$, and this one contains $B$. Of course $A+B/B$ is the set of cosets of the elements of $A$.

Answer 2

The error is if $B\not\subset A$ then we cannot take the quotient. This theorem essentially says if this is the case, then we can either expand $A$ to the smallest group containing $A$ and $B$ and then quotient by $B$, or we can restrict to $A$ and quotient by $A\cap B$ and, they give the same result.

Note that if $B\subset A$, then $A+B=A$ so the quotient is indeed $A/B$.

Answer 3

No, but it is "the same" (isomorphic to) as $A/(A \cap B)$, that is, modding by the part of $B$ that lies inside $A$. Explicitly, we send:

$(a+b) + B\ (= a + B) \mapsto a + (A\cap B)$.

Answer 4

The problem here is, since $B \not\subseteq A$, how to even make sense of $A/B$.

There are two reasonable ways:

• We can still make sense of the congruence relation $x \sim y$ when $x - y \in B$, and take the quotient of $A$ by this equivalence relation. This is the same thing as $A / (A \cap B)$.
• Since everything in $B$ should vanish anyways, adding them in shouldn't affect the result but gives a group that does contain $B$. This is the calculation $(A + B) / B$.

By the second isomorphism theorem, both approaches give isomorphic results.