Groups of order 2[superscript m] that contain cyclic subgroups of order 2[superscript m-3].
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Groups of order 2[superscript m] that contain cyclic subgroups of order 2[superscript m-3]. by Alice Madeleine McKelden

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Published in [n.p .
Written in English


Book details:

Edition Notes

[Reprinted from the American Mathematical Monthly, June-July, 1906]

The Physical Object
Pagination20 p.
Number of Pages20
ID Numbers
Open LibraryOL16961303M

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Notice we rarely add or subtract elements of \(\mathbb{Z}_n^*\). For one thing, the sum of two units might not be a unit. We performed addition in our proof of Fermat’s Theorem, but this can be avoided by using our proof of Euler’s Theorem did need addition to prove that \(\mathbb{Z}_n^*\) has a certain structure, but once this is done, we can focus on multiplication. Groups of order p [superscript m] which contain cyclic subgroups of order p [superscript m-3] Lewis Irving Neikirk This volume is produced from digital images from the Cornell University Library Historical Mathematics Monographs collection. Why is it the case that abelian group that contains a pair of cyclic subgroups of order 2 must contain a subgroup of order 4? Stack Exchange Network Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group. share | cite | improve this answer | follow |.

Order of a Cyclic Group and of its Elements, Gen-erators of a Cyclic Group The next theorem and corollaries are related to the order of ak and the groups generated by it. They will help us –nd generators of a cyclic group. Theorem (ak = agcd(n;k)) Let Gbe a group and a2Gsuch that jaj= n. Suppose further that k2Z+. Then: 1. ak. Full text of "Groups of order p [superscript m] which contain cyclic subgroups of order p [superscript m-3]" See other formats CORNELL UNIVERSITY LIBRARIES Mathematica Library White Hall 3 DATE DUE GAYLORD PRINTEDINU.S.A. Ui ^y. So the rst non-abelian group has order six (equal to D 3). One reason that cyclic groups are so important, is that any group Gcontains lots of cyclic groups, the subgroups generated by the ele-ments of G. On the other hand, cyclic groups are reasonably easy to understand. First an easy lemma about the order of an element. Lemma 3 contains exactly 2 elements of order 3. Further, no two cyclic subgroups of order 3 can contain the same element of order 3 since then they’d both contain that element, its inverse, and, of course, the identity. Thus, since the group has 8 elements of order 3, it must contain exactly 4 cyclic subgroups of order 3. By the previous problem.

Subsection Subgroups of Cyclic Groups. We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If \(G\) is a group, which subgroups of \(G\) are cyclic? If \(G\) is a cyclic group, what type of subgroups does \(G\) possess? Theorem Every subgroup of a cyclic group is cyclic. Proof. Finite cyclic groups. For every finite group G of order n, the following statements are equivalent. G is cyclic.; For every divisor d of n, G has at most one subgroup of order d.; If either (and thus both) are true, it follows that there exists exactly one subgroup of order d, for any divisor of statement is known by various names such as characterization by subgroups. have order 4, and so H = h[3]i is a cyclic subgroup of order 4. The element [9] = [3]2 has order 2. Since 92 ≡ 1 (mod 20), the subset K = {±[1],±[9]} is closed under multiplication. Since H is a finite, nonempty subset of a known group, Corollary implies that it is a subgroup. Since no element of H has order four, it is not cyclic.   Theorem. Let [math]G[/math] be a cyclic group of order [math]n[/math]. For each [math]d \mid n[/math], there exists a unique subgroup of order [math]d[/math]. Proof.