Theorem 3 14

Statement:

Suppose $\langle S, * \rangle$ has an identity element $e$ 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 $S'$.

Proof:

Let $s' \in S'$. We must show that $\phi (e)*'s'=s'*'\phi (e)=s'$. Because $\phi$ is an isomorphism, it is a one-to-one map of $S$ onto $S'$. In particular, there exists $s \in S$ such that $\phi (s)=s'$. Now $e$ is an identity element for $*$ so that we know that $e*s=s*e=s$. Because $\phi$ is a function, we then obtain

(1)
\begin{align} \phi (e*s)=\phi (s*e)=\phi (s). \end{align}

Using the definition of an isomorphism, we can rewrite this as

(2)
\begin{align} \phi (e)*'\phi (s)=\phi (s)*'\phi (e)=\phi (s). \end{align}

Remembering that we chose $s\in S$ such that $\phi (s)=s'$, we obtain the desired relation $\phi (e)*'s'=s'*'\phi (e)=s'$.

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