I need to prove that $(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime} \times G_{2}^{\prime}$, where $G^{\prime}$ denotes the commutator subgroup of $G$ – i.e., $G^{\prime}=[G,G]$, the subgroup generated by all commutators of elements of $G$.
Recall that the commutator of two elements $x$ and $y$ in a group $G$, denoted $[x,y]$ is defined to equal $x^{-1}y^{-1}xy$.
I have posted my proof as an answer below. Could somebody please take a look at it and let me know if it's okay? If not, please let me know what I need to do in order to fix it. Thank you. 🙂
Edit: I have been informed that my answer given below actually shows only equality of the set of commutators, and that $G^{\prime}$ consists of products of commutators. So, this question is no longer a proof check, but I am asking specifically how to fix what I have written below in order to make it actually answer the thing I set out to prove. I am having a bit of trouble understanding some of the hints given me thus far, and am going to need more detail in any answers given in order for them to be helpful.
Best Answer
Okay...I think this answer finally takes care of the difficulty. Thanks to arctictern for all his help :)
Let $g \in (G_{1}\times G_{2})^{\prime}$. Then, since $(G_{1} \times G_{2})^{\prime}$ is the subgroup generated by all commutators of elements of $G_{1} \times G_{2}$, $g$ is the product of commutators of elements of $G_{1} \times G_{2}$. Letting $(a_{i},b_{i})$, $(c_{i},d_{i})\in G_{1} \times G_{2}$, we have that
$\begin{align} g = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})]\,\text{for some}\, t, \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = ([a_{1},c_{1}]\cdot [a_{2},c_{2}]\cdot \,\, \cdots \, \, \cdot [a_{t},c_{t}], \,[b_{1},d_{1}]\cdot [b_{2},d_{2}]\cdot \,\, \cdots \, \, \cdot [b_{t},d_{t}]) \in G_{1}^{\prime} \times G_{2}^{\prime}\end{align}$
So, we have the inclusion $\mathbf{(G_{1}\times G_{2})^{\prime}\subseteq G_{1}^{\prime}\times G_{2}^{\prime}}$
In the other direction, let $h \in G_{1}^{\prime} \times G_{2}^{\prime}$.
To belong in $ G_{1}^{\prime} \times G_{2}^{\prime}$, $h$ must be the direct product of an element of $G_{1}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{1}$, and an element of $G_{2}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{2}$.
Let $h_{1}$ be such an element of $G_{1}^{\prime}$ and let $h_{2}$ be such an element of $G_{2}^{\prime}$, and let $h = h_{1} \times h_{2}$. Then,
$h_{1} = [a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}]$ for some $t$, where each $a_{i}$, $c_{i}$ is an element of $G_{1}$
and $h_{2} =[b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}]$ for some $t$, where each $b_{i}$, $d_{i}$ is an element of $G_{2}$.
So,
$\begin{align}h = h_{1} \times h_{2} = ([a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}], [b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}])\\ =(a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})] \in (G_{1}\times G_{2})^{\prime} \end{align}$
So, we have the inclusion $\mathbf{G_{1}^{\prime}\times G_{2}^{\prime} \subseteq (G_{1}\times G_{2})^{\prime}}$.
Thus, since we have inclusion in both directions, we have established the equality $\mathbf{(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime}\times G_{2}^{\prime}}$ for the right things this time, hopefully.