Question

Define a cyclic group. prove that a
group of order 3 is cyclic.​

Answer 1

$\mathfrak{\huge{\pink{Answer:-}}}$

$\bold{\pink{Cyclic\:group}}$

A cyclic group is a Group (mathematics) whose members or elements are powers of a given single (fixed) element , called the generator .

A finite cyclic group consisting of n elements is generated by one element , for example p, satisfying pn=Ipn=I, where II is the identity element .Every cyclic group is abelian .

Proof :-

You're trying to PROVE that GG is cyclic, so you cannot (yet) assert that x=gnx=gn. Instead, consider x⋅xx⋅x. It must be either x,y,x,y, or ee. If it's xx, then you have

x2=xx2(x−1)=xx−1x=e

x2=xx2(x−1)=xx−1x=e

which is a contradiction, because xx and ee are distinct elements of the group.

If x2=ex2=e, then xx has order 2, but 2 does not divide 3, so this contradicts Lagrange's theorem.

Finally, we conclude that x2=yx2=y, and thus the group is cyclic, generated by the element g=xg=x.

{Alternative if you don't like Lagrange yet:}

In the case where we suppose that x2=ex2=e:

The elements xe,xx,xe,xx, and xyxy must all be distinct for if two were the same, then multiplying by x−1x−1 on the left would show that two of e,x,ye,x,y were the same, which is impossible.

Since xe=xxe=x and we're assuming x2=ex2=e, we must have

xy=y.

xy=y.

multiplying on the right by y−1y−1 gives x=ex=e, a contradiction. So x2=ex2=e is also impossible.