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    If $A$ and $B$ are normal and $G/A$, $G/B$ are abelian then $G/(A \cap B)$ is abelian

    P-N4ABN

    Area
    Algebra
    Topics
    Abelian Groups, Normal Subgroups
    Status
    draft

    If \(A\) and \(B\) are normal in \(G\), and \(G/A\) and \(G/B\) are abelian, show that \(G/(A \cap B)\) is also abelian.

    Solution

    1. For a normal subgroup \(N\lhd G\), the quotient \(G/N\) is abelian if and only if \([G,G]\subseteq N\).

    \(G/N\) is abelian exactly when \(xyx^{-1}y^{-1}N=N\) for all \(x,y\in G\), that is, when every commutator lies in \(N\); since \(N\) is a subgroup, this is equivalent to \([G,G]\subseteq N\).

    2. \([G,G]\subseteq A\cap B\).

    Step 1 applied to \(A\) and to \(B\) gives \([G,G]\subseteq A\) and \([G,G]\subseteq B\).

    3. Q.E.D.

    The intersection \(A\cap B\) of normal subgroups is normal, so step 1 applied to \(N=A\cap B\), together with step 2, shows that \(G/(A\cap B)\) is abelian.

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