All groups of order 14

necessaryh 2021-02-10 Answered
All groups of order 14

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Answered 2021-02-11 Author has 4505 answers

Let G be the group of order 14. Claim is either \(\displaystyle{G}\stackrel{\sim}{=}\mathbb{Z}_{{{14}}}\), i.e., the cyclic group of order 14 or \(\displaystyle{G}\stackrel{\sim}{=}{D}_{{{14}}}\), i.e., the dihedral group of order 14. It is well known fact that a group of even order must contain an element of order 2, say a. Now by Lagrange theorem we know that order of an element of a finite group must divide its order. So, for any element order will be one of 1,2,7and 14. We wish to show that there is an element of order 7 and we will denote it by b.
case-1: Suppose there exist an element order of \(c=14\), So, we take \(\displaystyle{c}^{{2}}={b}\), therefore order of b is 7 and cc generates G, hence G is a cyclic group of order 14.
case-2: Suppose there does not exist any element of order 14 or 7. So, any element other than a,e will have order 2. Now let c not equal a is an element of order 2. Now by our assumption ac is an element of order 2. It can be shown that \(\{e,a,c,ac\}\) forms a subgroup of G, (since \(ac=ca\)). But this is impossible, since order of subgroup divides order of G. So, there exists an element of order 7, say b. Now consider \(\displaystyle{H}={\left\lbrace{e},{b},{b}^{{2}},\ldots,{b}^{{6}}\right\rbrace}\) is subgroup of order 7. Since aa has order 2 so, aa not in H, because all the elements of H are of order 7. Therefore, H and aHaH two distinct disjoint cosets of H, each of them consists of 7 elements. Therefore,
Now consider the element ba, it must be one of the ab^j, where \(\displaystyle{j}\in{\left\lbrace{1},{2},⋯ ,{6}\right\rbrace}\). Suppose \(ba=ab\), then G will be abelian, since G is collection of elements like \(\displaystyle{a}^{{i}}{b}^{{k}}\), where \(\displaystyle{0}\le{i}\le{1}\) and \(\displaystyle{0}\le{k}\le{6}\) and order of abab will be 14, hence G will be cyclic. Otherwise, Since by sylow theorem \(\displaystyle{G}{G}\) have only one subgroup of order 7 and that is H, so, order of baba must be 2. ( Order of baba can not be 7, since baba not in H and order of baba can not be 14, otherwise G is cyclic and hence abelian, which is not the case.).Therefore,
\(\displaystyle{b}{a}={\left({b}{a}\right)}^{{-{1}}}={a}^{{-{1}}}{b}^{{-{1}}}-{a}{b}^{{6}}{\left({sin{{c}}}{e}\ {a}^{{2}}={b}^{{7}}={e}\right)}\)
\(\displaystyle{G}={\left\lbrace{a}^{{i}}{b}^{{k}}:{w}{h}{e}{r}{e}\ {0}\le{i}\le{1}{\quad\text{and}\quad}{0}\le{k}\le{6},{b}{a}={a}{b}^{{6}}\right\rbrace}\), that is it is isomorphic to \(\displaystyle{D}_{{{14}}}\)

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