# Show that the identity maps to the identity

1. Feb 19, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Suppose that $\langle S,*\rangle$ has an identity e for *. If $\phi : S \rightarrow S'$ is an isomorphism of $\langle S,*\rangle$ with $\langle S',*\rangle$, then $\phi (e)$ is an identity element for the binary operation $*'$ on S'.

2. Relevant equations

3. The attempt at a solution
We know that $e*s = s*e = s$. Since $\phi$ is a function, we then have $\phi (e*s) = \phi (s*e) = \phi(s)$. By definition of homomorphism, we have that $\phi (e) *' \phi (s) = \phi (s) *' \phi (e) = \phi(s)$. Since $\phi$ is a surjection, we know that there exists an s in S s.t. for all s' in S' $\phi(s) = s'$. Thus $\phi (e) *' s' = s' *' \phi (e) = s'$, for all s' in S'.

This seems to be the correct proof. However, nowhere do I use the fact that phi is injective. Does this mean that I have done something wrong?

2. Feb 19, 2017

### Staff: Mentor

The last sentence is a bit weird, because the for all quantifier is misplaced. It should be $\forall \; s' \in S' \, \, \exists \; s\in S \, : \, \phi (s)=s'$ not the other way around. These quantifiers don't commute. I have the suspicion that it is because you did the second step first.

You want to show that $\phi (e)$ is an identity element in $S'$. So you have to show that $\phi (e) * s' = s'$ for an arbitrary element $s' \in S'$. Now you use the surjection to conclude, that there is an element $s \in S$ with $\phi (s) =s'$ and apply your first step. You switched the order.

$s'$ is chosen arbitrary and $s$ is not, as it has to map to the special choice $s'$. It exists due to surjectivity, but the for all quantifier isn't used correctly. Arbitrary doesn't mean for all. You show it for one, and at the end, it holds for all as long as you don't pose any restrictions on the choice of $s'$. So you probably meant the right thing, but didn't phrase it the right way. If you start with for all $s'$ instead of an arbitrary but fixed $s'$ - which you can do - then you should write $s=s(s')$ as it depends on $s'$ and is not the same for all $s'$. But this is harder to read than a single example $s'$ where we state afterwards, that it is the same procedure for all and thus hold for all.
Surjectivity (along with homomorphy which you didn't mention in the beginning) is sufficient. In fact the entire kernel of $\phi$ maps to $1' \in S'$.