Every quotient of an abelian group is abelian

Let A be an abelian group and let B \leq A. Prove that A/B is abelian. Give an example of a nonabelian group G containing a proper normal subgroup N such that G/N is abelian.

Lemma 1: Let G be a group. If g \in G such that g^2 = g, then g = 1. Proof: Use cancellation. \square

Lemma 2: Let G be a group. If |G| = 2, then G \cong Z_2. Proof: Since G = \{ e a \} has an identity element, say e, we know that ee = e, ea = a, and ae = a. If aa = a, we have a = e, a contradiction. Thus aa = e. We can easily see that G \cong Z_2.

If A is abelian, every subgroup of A is normal; in particular, B is normal, so A/B is a group. Now let xB, yB \in A/B. Then (xB)(yB) = (xy)B = (yx)B = (yB)(xB). Hence A/B is abelian.

Consider \langle i \rangle \leq Q_8. We saw in a previous exercise that the normalizer of \langle i \rangle in Q_8 is all of Q_8, so that \langle i \rangle is normal. Since Q_8 has 8 elements and \langle i \rangle has 4 elements, Q_8/\langle i \rangle has 2 elements. Thus Q_8/\langle i \rangle \cong Z_2, which is abelian.

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