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Abel's theorem in problems and solutions - School of Mathematics

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Solutions 141<br />

is commutative. S<strong>in</strong>ce is a normal subgroup <strong>of</strong> the<br />

group G (see 131)‚ we may consider the quotient group<br />

We prove that the subgroup is commutative. Consider the<br />

natural homomorphism (see §1.13) with kernel<br />

S<strong>in</strong>ce is surjective (see 155) Thus<br />

etc.. Consequently we have (because<br />

is the kernel <strong>of</strong> the homomorphism S<strong>in</strong>ce<br />

the subgroup (see 117) is commutative.<br />

Denote by the quotient group As before‚<br />

we can prove that the subgroup is commutative. Put<br />

etc.. The group is commutative<br />

(see 129). The sequence <strong>of</strong> groups with normal subgroups<br />

is the requested sequence.<br />

171. Let be a sequence <strong>of</strong> groups with the properties<br />

mentioned <strong>in</strong> Problem 170. We will prove that all the groups <strong>of</strong> the sequence‚<br />

<strong>in</strong> particular are soluble. We shall use the <strong>in</strong>duction from to<br />

0. The group is soluble‚ because by hypothesis it is commutative <strong>and</strong><br />

therefore Suppose that we have already proved that the<br />

group is also soluble. We will prove that the group too‚ is soluble.<br />

The group conta<strong>in</strong>s by hypothesis a normal‚ commutative‚ <strong>and</strong> thus<br />

soluble‚ subgroup from which the quotient group is<br />

soluble by the <strong>in</strong>duction hypothesis. Hence the group is also soluble<br />

by virtue <strong>of</strong> the result <strong>of</strong> Problem 166. In conclusion we can state that all<br />

the groups <strong>in</strong> the sequence are soluble‚ <strong>and</strong>‚ <strong>in</strong> particular‚<br />

is soluble.<br />

172. Any permutation <strong>of</strong> order can be written <strong>in</strong> the form<br />

where the are all dist<strong>in</strong>ct <strong>and</strong> take values from 1 to In particular‚<br />

may take any one <strong>of</strong> the the values. Later can take any one <strong>of</strong> the<br />

rema<strong>in</strong><strong>in</strong>g values‚ <strong>and</strong> so on. It follows that the number <strong>of</strong> dist<strong>in</strong>ct<br />

permutations <strong>of</strong> degree is equal to<br />

173. If <strong>and</strong><br />

then <strong>and</strong>

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