Given $\phi:G\to G_1$ a homomorphism, prove $\ker(\phi)=G$ I have the following problem

Let $G$ and $G_1$ be finite groups such that $\gcd(|G|,|G_1|)=1$, and let $\phi:G\to G_1$ be a group homomorphism. Prove that $ker \phi=G$.

We can assume $\ker\phi$ is a subgroup of $G$.
My attempted proof is as follows:

It would be highly appreciated if someone could help to prove this as I am preparing for a group theory test. Thanks :)
 A: Here is a proof using only basic facts.
Take any $x\in G$, set $y=\phi(x)$.  We have $x^{|G|} = e_G$, and additionally $y^{|G_1|} = e_{G_1}$, by a simple form of Lagrange.  Applying $\phi$ to the first equation, we have $y^{|G|} = y^{|G_1|} = e_{G_1}$.
Since $|G|$ and $|G_1|$ are relatively prime, there are integers $a,b$ such that $a|G| + b|G_1| = 1$.  Then $y = y^{a|G| + b|G_1|} = (y^{|G|})^a (y^{|G_1|})^b = e_{G_1}^a e_{G_1}^b = e_{G_1}$, i.e. $x\in\ker\phi$.
A: First a comment on your proof. $\phi(H)\subset H$ is not true. We have $\phi(H)\le G_1$, that is the image of $H$ is a subgroup of $G_1$, and unless $G\cap G_1\supset H$, we can't have $H\le G_1$. (But since the orders of $G$ and $G_1$ are coprime this cannot occur). 
I will give a simple proof of this using the first isomorphism theorem. Notice that $\phi(G)\cong G/\ker\phi$. Thus, $|\phi(G)||\ker\phi|=|G|$. But $\phi(G)\le G_1$, so the order of $\phi(G)$ divides the order of $G_1$. Thus, $|\phi(G)|$ divides both $|G_1|$ and $|G|$. Since these are coprime, it follows that $|\phi(G)|=1$, and thus $|\ker(\phi)|=|G|$, so $\ker\phi=G$. 
A: You should know (or can verify) that if $\phi$ is a group homomorphism, then $|\phi(x)|$ divides $|x|$. Hence, for all $x$, $\phi(x)$ must be chosen to fit this qualification. But then $|\phi(x)|$ divides both $|G|$ and $|G_1|$, so $|\phi(x)|=1$, so $\phi(x)$ is the group identity in $G_1$.
