I'm trying to understand the steps taken in this proof I saw in my group theory course.
It's a prof of the statement that if;
$\theta:G_1\rightarrow G_2$ is an isomorphism then , $\theta(e_1)=e_2.$ and goes as follows
$\theta(e_1)=\theta(e_1e_1)=\theta(e_1)\theta(e_1)$
Then we multiply through by $[\theta(e_1)]^{-1}$
To get
$\theta(e_1)[\theta(e_1)]^{-1}=\theta(e_1)\theta(e_1)[\theta(e_1)]^{-1}$
which gives us $e_2=\theta(e_1)$.
I don't really see how it does though as forgetting about groups and identities for a moment and just considering what I know about function and inverses I would have read the line before the statement is proved as
$f(f^{-1}(y))=f(f(f^{-1}(y))$
$f(x)=f(f(x))$
$f(x)=f(y)$
which doesn't really make any sense. What am I considering incorrectly here ?
Note: an edit was made to this post to fix an error that @Mike noticed
I am assuming $e_1$ is the identity element in $G_1$ and $e_2$ the identity element in $G_2$.
Shouldn't it be
$$\theta(e_1) = \theta(e_1e_1) = \theta(e_1)\theta(e_1)$$
(Check to make sure you see that this string of equations is indeed valid)
And then multiplying both sides by $[\theta(e_1)]^{-1}$, the LHS is $e_2$ (because $bb^{-1} = e_2$ for every $b \in G_2$, and $\theta(e_1)$ is indeed in $G_2$). While the RHS is $\theta(e_1)\theta(e_1)[\theta(e_1)]^{-1} =\theta(e_1)$.
This gives $e_2 = \theta(e_1)$, which is what you wanted.