All groups of order 14

necessaryh

necessaryh

Answered question

2021-02-10

All groups of order 14

Answer & Explanation

Khribechy

Khribechy

Skilled2021-02-11Added 100 answers

Let G represent the order 14 group. Claim is either G=Z14, i.e., the cyclic group of order 14 or G=D14, 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, we are aware that the order of an element in a finite group must be divided by the Lagrange theorem. Therefore, the order will be one of 1, 2, 7, or 14. We will use the symbol b to indicate the element of order 7 that we want to demonstrate.
case-1: Suppose there exist an element order of c=14, So, we take c2=b, therefore order of b is 7 and cc generates G, hence G is a cyclic group of order 14.
case-2: Assume that neither an element of order 14 nor 7 exists. Therefore, every element except for a,e will have order 2. Let c now differ from a, which is a component of order 2. Now, according to our presumption, AC is a member of order 2. It is demonstrable 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 H={e,b,b2,,b6} 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,
G={e,b,b2,,a,ab,ab2,ab6}.
Now consider the element ba, it must be one of the ab^j, where j{1,2,,6}. Suppose ba=ab, then G will be abelian, since G is collection of elements like aibk, where 0i1 and 0k6 and order of abab will be 14, hence G will be cyclic. Otherwise, Since by sylow theorem GG 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,
ba=(ba)1=a1b1ab6(since a2=b7=e)
Therefore,
G={aibk:where 0i1 and 0k6,ba=ab6}, that is it is isomorphic to D14

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