Groups whose orders have no common factors

[SNOOTCHIEBOOCHEE]

Homework Statement

Let G and G' be finite groups whose orders have no common factor. Prove that the homomorphism $$\varphi$$ G $$\rightarrow$$ G' is the trivial one $$\varphi$$ (x) =1 for all x.

The Attempt at a Solution

My thoughts are that we need to use lagrange's thm. somehow. or maybe not.

We have the order of G and G' such that. GCD( |G|, |G'|) =1.

$$\varphi$$ G $$\rightarrow$$ G'

Let |G| = n

By legranges thm.

gⁿ $$\in$$ G = 1_G

and we know that $$\varphi$$ (gⁿ)= 1_G'

But i don't really no what to do from here. It doesn't seem as if i am on the right track.

Any thoughts?

 

[HallsofIvy]

I presume you mean prove that the only homomorphism from G to G' is $\phi(g)= 1_{G'}$. Let x be a member of G, not equal to 1_G. Let n be the order of x: the smallest integer n such that xⁿ= 1_G. What is the order of $\phi(x)$?

 

[SNOOTCHIEBOOCHEE]

The order of $$\varphi(x)$$ must divide n.

 

[HallsofIvy]

But the order of any member of member of G must divide |G|. Since the order $\phi(x)$ divides n, it also divides |G|. And the order of $\phi(x)$ must, like the order of any member of G', must divide |G'|. What is the only number that divides both |G| and |G'|?

 

[SNOOTCHIEBOOCHEE]

Holy crap youre my hero.