I have proven cyclic group to be a group, but now must prove cyclic group order n to be isomorphic to Zn. I realize I must show homomorphism, injection and surjection. But not struggling with initial equivalences. Please help.(Crying)

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- Nov 20th 2011, 03:29 PMMAdoneProving isomorphism to Zn
I have proven cyclic group to be a group, but now must prove cyclic group order n to be isomorphic to Zn. I realize I must show homomorphism, injection and surjection. But not struggling with initial equivalences. Please help.(Crying)

- Nov 20th 2011, 03:35 PMJose27Re: Proving isomorphism to Zn
Let $\displaystyle a$ be a generator of your cyclic group $\displaystyle G$, can you find a function such that $\displaystyle f:G \to \mathbb{Z}_n$ and $\displaystyle f(a)=1$ such that $\displaystyle f$ is an homomorphism?

- Nov 20th 2011, 03:46 PMMAdoneRe: Proving isomorphism to Zn
are you suggesting a concrete f like f(x)=x+5 or generalized like f(x)=ix+j.

- Nov 20th 2011, 03:49 PMTheChazRe: Proving isomorphism to Zn
Something concrete, in terms of things we know must exist (like generator "a").

Try a^n - Nov 20th 2011, 03:53 PMJose27Re: Proving isomorphism to Zn
????

Suppose the operation of your group is given by $\displaystyle +$, if an $\displaystyle f$ as I described existed, what would be the image of $\displaystyle ka=a+a+a\hdots a$ ($\displaystyle k$ summands) under $\displaystyle f$, and $\displaystyle a$ is a generator so... - Nov 20th 2011, 04:39 PMMAdoneRe: Proving isomorphism to Zn
all elements of the cyclic group

- Nov 20th 2011, 08:04 PMDevenoRe: Proving isomorphism to Zn
MAdone, this is really the whole of the idea.

if $\displaystyle f:G \to \mathbb{Z}_n$ is to be a homomorphism, we must have f(a*a) = f(a) + f(a) = 1+1. so there is really only "one" essential way to define f:

$\displaystyle f(a^k) = k$. that f is a homomorphism follows immediately from the laws of exponents.

since G is finite it suffices to show that f is surjective (clearly |G| = $\displaystyle |\mathbb{Z}_n|$ = n), and thus f is an isomorphism.