# All groups of order 14

All groups of order 14

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Khribechy

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,
$$\displaystyle{G}={\left\lbrace{e},{b},{b}^{{2}},\ldots,{a},{a}{b},{a}{b}^{{2}},\ldots{a}{b}^{{6}}\right\rbrace}.$$
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,
$$\displaystyle{b}{a}={\left({b}{a}\right)}^{{-{1}}}={a}^{{-{1}}}{b}^{{-{1}}}-{a}{b}^{{6}}{\left({sin{{c}}}{e}\ {a}^{{2}}={b}^{{7}}={e}\right)}$$
Therefore,
$$\displaystyle{G}={\left\lbrace{a}^{{i}}{b}^{{k}}:{w}{h}{e}{r}{e}\ {0}\le{i}\le{1}{\quad\text{and}\quad}{0}\le{k}\le{6},{b}{a}={a}{b}^{{6}}\right\rbrace}$$, that is it is isomorphic to $$\displaystyle{D}_{{{14}}}$$