# Thread: Prove f is a group morphism and establish 1 condition

1. ## 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

2. 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
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).

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

3. 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|).