Search the corpus

    If $N$ and $G/N$ are solvable then $G$ is solvable

    P-S2KMD

    Area
    Algebra
    Topics
    Normal Subgroups, Solvable Groups
    Status
    draft

    Let \(N\normal G\). Show that if \(N\) and \(G/N\) are solvable, then \(G\) is solvable.

    Solution

    Let \(\pi:G\to G/N\) be the quotient map. Since \(G/N\) is solvable, its derived series reaches the identity: \[ (G/N)^{(r)}=1 \] for some \(r\). Quotient maps preserve commutators, so \[ \pi(G^{(r)})=(G/N)^{(r)}=1. \] Hence \[ G^{(r)}\le N. \]

    Since \(N\) is solvable, there is some \(s\) with \(N^{(s)}=1\). Because derived subgroups are monotone under inclusion, \[ G^{(r+s)}=(G^{(r)})^{(s)}\le N^{(s)}=1. \] Thus the derived series of \(G\) terminates, so \(G\) is solvable.

    Source Collections