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The definition for a solvable group I am using (from University lecture notes) is:

A group $G$ is solvable if we can find a sequence of $n$ subgroups: $$\{1\}=G_0 \trianglelefteq G_1\trianglelefteq\ldots\trianglelefteq G_{n-1} \trianglelefteq G_n = G$$

Where $G_{i-1}$ is a normal subgroup of $G_i$ and $G_i / G_{i-1}$ is abelian.

This has motivation from the Galois correspondence of a solvable field extension, I recognise that the sequence of normal subgroups corresponds to a sequence of normal extensions of $\mathbb{Q}$ leading up to the corresponding Galois extension of G, but I am struggling to see what the requirement of abelian factor groups correspond to in the field extensions? Is this something relating to radical extensions or am I missing a part of the definition?

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    $\begingroup$ Finite solvable groups are precisely the Galois groups of finite field extensions that can be obtained by radical extension, that is, extensions $F(\alpha)/F$ where $\alpha$ can be obtained in terms of sums, products, quotients, and radicals of elements of $F$. This is of course connected to the problem of solving a polynomial equation by radicals. There's some extra amount of work needed (contained in what is sometimes called "Kummer Theory"), plus the fact that you can express any finite abelian group as a product of cyclic groups. $\endgroup$
    – Arturo Magidin
    Feb 22 at 19:09
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    $\begingroup$ See also radical extensions. $\endgroup$
    – Arturo Magidin
    Feb 22 at 19:14
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    $\begingroup$ Does this help you? math.stackexchange.com/questions/3753328/… $\endgroup$
    – Vercassivelaunos
    Feb 22 at 19:14
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    $\begingroup$ There's an equivalent definition of being solvable, which demands that each quotient $G_i/G_{i-1}$ is cyclic rather than just abelian. I think this definition makes the connection to radical extensions much clearer. If $L/K$ is solvable by radicals, then we can write $L/K$ as a tower of cyclic extensions: first add enough prime power roots of unity to get started; then, extensions of the form $K'(\zeta_n, \sqrt[n]a)/K'(\zeta_n)$ have cyclic Galois group. Conversely, by Kummer theory, all cyclic order $n$ extensions of $K'(\zeta_n)$ are of the form $K'(\zeta_n, \sqrt[n]a)$ for some $a$. $\endgroup$
    – Mathmo123
    Feb 22 at 19:50
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    $\begingroup$ @Mathmo123: Note of course that the definition with cyclic factors is only equivalent to the definition with abelian factors in the finite case. $\endgroup$
    – Arturo Magidin
    Feb 22 at 21:52

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