Finding a subgroup of $S_8$ isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_4$ I've resorted to using the fact that $\mathbb{Z}_4 \times \mathbb{Z}_4$ has $16$ elements and if you check there are going to be $10$ of order $4$ and $2$ of order two.
Therefore I've just let $H \leq G$ a subgroup of $S_8$ which has two elements of order $2$ and $10$ or order $4$. For example
$$
H = \{(1),(12),(13),(1234),(1324),(2345),(2435),(3456),(3546),(4567),(4657),(5678),(5768)\}
$$
Is this approach alright, and what might be a more clever way to go about solving this?
Thanks in advance.
 A: 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: $H=\{a,a^2,a^3,a^4,b,b^2,b^3,ab,ab^2,ab^3,a^2b,a^2 b^2,a^2b^3,a^3b,a^3b^2,a^3b^3\}$
A: 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
$$\begin{align}
A'B'&=\{ab\in S_8\mid a\in A', b\in B'\}\\
&=\{(\color{red}{e},e), (\color{red}{e}, (5678)), (\color{red}{e},(5678)^2),(\color{red}{e},(5678)^3)\\
&(\color{green}{(1234)}, e), (\color{green}{(1234)}, (5678)), (\color{green}{(1234)},(5678)^2),(\color{green}{(1234)},(5678)^3)\\
&(\color{blue}{(1234)^2},e), (\color{blue}{(1234)^2}, (5678)), (\color{blue}{(1234)^2},(5678)^2),(\color{blue}{(1234)^2},(5678)^3)\\
&(\color{pink}{(1234)^3},e), (\color{pink}{(1234)^3}, (5678)), (\color{pink}{(1234)^3},(5678)^2),(\color{pink}{(1234)^3},(5678)^3)\}\\
&=A\times B\\
\end{align}$$
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)".
