Prove f is a group morphism and establish 1 condition

Please can you help me with this question:

Define the following map on a cyclic finite group G:

f:G--> G x|-->f(x):=x^m

where m is a fixed interger prove f is a group morphism and establish one condition whether sufficent of necessary for f to be a monomorphism.

I know that x^m is a cyclic group

I also know a monomorphism is an injective homomorphism, which is one to one as a function which relates maybe to f:G-->G.

When it asks to prove it is a group morphism does this mean I need to find the kernal and the image.

I know what the question means but just don't know the proof of it.

Any help would be great thanks

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Originally Posted by imagenius

Please can you help me with this question:

Define the following map on a cyclic finite group G:

f:G--> G x|-->f(x):=x^m

where m is a fixed interger prove f is a group morphism and establish one condition whether sufficent of necessary for f to be a monomorphism.

I know that x^m is a cyclic group

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Technically, x^m is a member of the group. The set of all powers of x, {x^m}, is a cyclic group. And, since you are given that G is cyclic, it follows that {x^m} is a subgroup of G having index(x) members. Now, think about factors of ord(G).

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I also know a monomorphism is an injective homomorphism, which is one to one as a function which relates maybe to f:G-->G.

When it asks to prove it is a group morphism does this mean I need to find the kernal and the image.

I know what the question means but just don't know the proof of it.

Any help would be great thanks

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no, to prove it is a group morphism, you need to show that f preseves the multiplication.

since G is finite, a monomorphism will be an isomorphism.

for example, if m = |G|, it should be clear x-->x^m will NOT be a monomorphism

(since everything gets mapped to e). hint: consider gcd(m, |G|).