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Every quotient of an abelian group is abelian


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.3

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.


Solution:

Lemma: 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 $a^2 = a$, we have $a = e$, a contradiction. Thus $a^2 = e$. We can easily see that $G \cong Z_2$. $\square$

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.


Linearity

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