Question: Let $A$, $B$, $C$ be groups. Assume that $A \cong B$ and $B\cong C$. Prove that $A \cong C$.
Proof:
Let $\phi: A \to B$ and $\sigma: B \to C$
Suppose $\sigma \circ \phi: A \to C$
$i). One-to-one:$
Then $\sigma \circ \phi$ is one-to-one. Let $a, b \in A$
Suppose $(\sigma \circ \phi)(a) = (\sigma \circ \phi)(b)$
$\Rightarrow$ $\sigma (\phi(a)) = \sigma (\phi(b))$
(Since $\sigma$ is one-to-one.)
$\Rightarrow$ $(\phi(a)) = (\phi(b))$
(Since $\phi$ is one-to-one.)
$\Rightarrow$ $a = b$
$ii). Onto:$
Then $\sigma \circ \phi$ is onto.
Let $c \in C$. Because $\sigma$ is onto there exists $b \in B$ such that $\sigma(b) = c$. Also since $\phi$ is onto there exists a $a \in A$ such that $\phi(a) = b$.
Then $\sigma(\phi(a)) = \sigma(b) = c$
Then $\sigma \circ \phi$ is onto.
$iii).Opreation - Perserving:$
Let $x_1, x_2 \in A$.
Then $(\sigma \circ \phi) = \sigma(\phi(x_1, x_2)) = \sigma(\phi(x_1)\phi(x_2))$
Because $\phi(x_1) \in B$ and $\phi(x_2) \in B$,
$\sigma(\phi(x_1)\sigma(\phi(x_2)) = (\sigma \circ \phi)(x_1)(\sigma \circ \phi)(x_2)$
Then $\sigma \circ \phi$ is operation-preserving.
$\therefore A \cong C$
I know it isn't the most elegant proof, but I was wondering if this is totally correct or if I missed something or did something incorrectly. Any help appreciated!