Basic Ring Theory Question involving the unit element

Question

If $ \phi: R \to R'$ is a homomorphism of $R$ onto $R'$ and $R$ has a unit element, $1$, show that $\phi(1)$ is the unit element of $R'$.

I am having trouble proving this. I think I just need a hint on how to start and I will be able to solve. But basically I have tried a few things like $\phi(1) = \phi(a)\phi(b)$ but I cannot assume it is a division ring. Also is it okay to assume from the wording of the question that $R'$ has a unit element.

Thanks in advance.

Answer 1

Take any $r' \in R$ and show that both $\phi(1) r' = r'$ and $r' \phi(1) = r'$. How to do this?

Hint: the homomorphism $\phi$ is onto. (Mouse over box to reveal the answer.)

Since $\phi$ is onto, there is some $r \in R$ such that $\phi(r) = r'$. Now, $$ \phi(1) r' = \phi(1) \phi(r) = \phi(1r) = \phi(r) = r'. $$ Multiplication on the right by $\phi(1)$ is completely analogous.

Answer 2

$\phi(a)=\phi(1\cdot a)=\phi(1)\cdot \phi (a)$ where $a\in R$

Answer 3

Since $\phi$ is onto, we have $1'=\phi(r)$ for some $r\in R$. But then $1'=\phi(r)=\phi(r\cdot1)=\phi(r)\phi(1)$, as needed.