Finding a subgroup of $S_8$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_4$

Well, Consider $S_8$.

Take, $a=(1234) $ , $b=(5678) $

We know order of a cycle is the length of the cycle. Hence, $|a|=|b|=4$

Again we know any two disjoint cycles commute.

Now consider,$H=\langle a, b : a^4 =b^4 =id,ab=ba\rangle $

Claim : $H\cong \mathbb{Z_4 ×Z_4}$

Hint: Send generators to generators .

Edit: $H=\{a,a^2,a^3,a^4,b,b^2,b^3,ab,ab^2,ab^3,a^2b,a^2 b^2,a^2b^3,a^3b,a^3b^2,a^3b^3\}$

Let $A=\langle (1234)\rangle, B=\langle(5678)\rangle.$ Then $|A|=|B|=4$ and $A\cong B\cong \Bbb Z_4$.

Since $\Bbb Z_4\times \Bbb Z_4$ is abelian, each of its subgroups is normal. In particular, $A'=A\times \{e\}\cong \Bbb Z_4\times \{e\}\unlhd \Bbb Z_4\times \Bbb Z_4$ and $B'=\{e\}\times B\cong \{e\}\times \Bbb Z_4\unlhd \Bbb Z_4\times \Bbb Z_4$.

Clearly $A'\cap B'$ is trivial and $A'$ commutes with $B'$.

Consider

$$\begin{align} A'B'&=\{ab\in\mid a\in A', b\in B'\}\\ &=\{(\color{red}{e},e), (\color{red}{e}, (5678)), (\color{red}{e},(5678)^2),(\color{red}{e},(5678)^3)\\ &(\color{green}{(1234)}, e), (\color{green}{(1234)}, (5678)), (\color{green}{(1234)},(5678)^2),(\color{green}{(1234)},(5678)^3)\\ &(\color{blue}{(1234)^2},e), (\color{blue}{(1234)^2}, (5678)), (\color{blue}{(1234)^2},(5678)^2),(\color{blue}{(1234)^2},(5678)^3)\\ &(\color{pink}{(1234)^3},e), (\color{pink}{(1234)^3}, (5678)), (\color{pink}{(1234)^3},(5678)^2),(\color{pink}{(1234)^3},(5678)^3)\}\\ &=A\times B\\ \end{align}$$

Thus $A\times B$, the interior direct product of $A'$ and $B'$, is isomorphic to $\Bbb Z_4\times \Bbb Z_4$.

For more details on the difference between an internal direct product and an external one, see Gallian's "Contemporary Abstract Algebra (Eighth Edition)".