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Let

$$ \phi : \begin{array}{c} 1 \mapsto 2 \\ 2 \mapsto 1 \\ 3 \mapsto 3 \\ \end{array} \qquad \text{and} \qquad \psi : \begin{array}{c} 1 \mapsto 2 \\ 2 \mapsto 3 \\ 3 \mapsto 1 \\ \end{array}. $$

Define the group $S_3 = \{e, \phi, \psi, \psi^2, \phi\psi, \phi\psi^2\}$ whose operation is composition of functions: for $a,b \in S_3$, take $(ab)(x) = (b \circ a)(x) = b(a(x))$.

Let $f : S_3 \to \langle\phi\rangle$ be defined by $f(\phi^i \psi^j) = \phi^i$. Example 2.7.2 of Herstein's Topics in Algebra challenges the reader to show that $f$ is a homomorphism. I did so by making a table and checking all 36 cases. Is there a less tedious way?

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You should note that defining homomorphisms by $f(\phi^i\psi^j)=\phi^i$ requires that we know that all elements of $S_3$ can be uniquely represented as $\phi^i\psi^j$ for exactly one pair $i,j$. (Keep in mind when working with other groups that defining homomorphisms like this is, in general, bad practice and can get you in trouble - in other words, you may find yourself working with homomorphisms that are not well-defined. Proving this uniqueness is absolutely necessary.) Of course, the author did it here, so we will take for granted that it is well defined.

With that noted, my hint is to look at the form of the permutations $\psi^j$ when switched with $\phi^i$ for each case of $i$ - in other words, determine what $\phi^{-i}\psi^{j}\phi^{i}$ is equal to. Then write out the homomorphic property proof explicitly using this identity on both sides.

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Consider the composition $S_3 \to S_3 / \langle \psi \rangle \cong \langle \phi \rangle$. You can check that it is exactly $f$.

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