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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.

2 Answers 2

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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.

3

Let $(g_{1},g_{2})\,,\,(h_{1},h_{2}) \in G_{1}\times G_{2}$. Then, $\begin{align}[(g_{1},g_{2})(h_{1},h_{2})]=(g_{1},g_{2})^{-1}(h_{1},h_{2})^{-1}(g_{1},g_{2})(h_{1},h_{2}) \\ =(g_{1}^{-1},g_{2}^{-1})(h_{1}^{-1},h_{2}^{-1})(g_{1},g_{2})(h_{1},h_{2}) = (g_{1}^{-1}h_{1}^{-1}g_{1}h_{1},g_{2}^{-1}h_{2}^{-1}g_{2}h_{2}) \\ = ([g_{1},h_{1}],[g_{2},h_{2}]) = ([g_{1},h_{1}],e_{G_{2}})(e_{G_{1}},[g_{2},h_{2}])\end{align}$.

So, each commutator in $G_{1} \times G_{2}$ is the product of one in $G_{1}$ with one in $G_{2}$.

Therefore, $\mathbf{(G_{1}\times G_{2})^{\prime}\subseteq G_{1}^{\prime}\times G_{2}^{\prime}}$.

Now, to get the inclusion in the other direction, let $(g,h) \in G_{1}^{\prime} \times G_{2}^{\prime}$. Then, $g = [g_{1},g_{2}]\in G_{1}^{\prime}$ and $h = [h_{1},h_{2}] \in G_{2}^{\prime}$.

So,

$\begin{align}(g,h) = ([g_{1},g_{2}],[h_{1},h_{2}]) = (g_{1}^{-1}g_{2}^{-1}g_{1}g_{2}, h_{1}^{-1}h_{2}^{-1}h_{1}h_{2}) \\ = (g_{1}^{-1},h_{1}^{-1})(g_{2}^{-1},h_{2}^{-1})(g_{1},h_{1})(g_{2},h_{2}) \\ = (g_{1},h_{1})^{-1}(g_{2},h_{2})^{-1}(g_{1},h_{1})(g_{2},h_{2}) \\= [(g_{1},h_{1}),(g_{2},h_{2})] \end{align} $

where $(g_{1},h_{1}), (g_{2},h_{2}) \in G_{1} \times G_{2}$.

Therefore, the product of a commutator in $G_{1}$ with a commutator in $G_{2}$ is a commutator in $G_{1} \times G_{2}$. Thus, $\mathbf{G_{1}^{\prime} \times G_{2}^{\prime} \subseteq (G_{1} \times G_{2})^{\prime}}$.

And so, we get our equality, $G_{1}^{\prime} \times G_{2}^{\prime} = (G_{1} \times G_{2})^{\prime}$.

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    Technically you've shown an equality of the *sets* of commutators. Remember elements of $G'$ will in general be products of commutators. – 2017-01-03
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    @arctictern so what more do I need to say to make it perfect? – 2017-01-03
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    For example, if you want to do it elementarily... Let $g\in(G_1\times G_2)'$. Then it can be written as $$g=[(a_1,b_1),(c_1,d_1)]\cdots[(a_t,b_t),(c_t,d_t)].$$ But this equals $$([a_1,c_1]\cdots[a_t,c_t],[b_1,d_1]\cdots[b_t,d_t]) $$ which is in $G_1'\times G_2'$. – 2017-01-03
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    @arctictern do you need to do induction for that? – 2017-01-03
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    In other words: do I need to write down an induction proof for you to understand why the equalities are true? Well, that's up to you. – 2017-01-03
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    @arctictern no I mean is an induction proof necessary to be written down in order to make the proof airtight and complete? – 2017-01-03
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    @arctictern also, do I need to begin using that notation from the very beginning of the proof? – 2017-01-03
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    I assume you're doing this for homework, in which case how you want to write up the proof is up to you. – 2017-01-03
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    @arctictern ack this is so frustrating! I just want to write it correctly! And I'm having trouble understanding what you're saying about the elements of $G^{\prime}$ being products of commutators. (Probably elementary, but how do you know that? I know they're commutators, but why products?) And if that's true, then what is the correct way to approach the proof? Do it the way I did in my answer and at the end say really quickly what you said there? Or, show what I showed but for elements in the form you wrote there.? – 2017-01-03
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    Do *you* understand the equality I wrote? Personally, I wouldn't bother writing down an induction proof since to me the equality is obvious, but whether you should depends on whether or not you think your instructor would expect it. I showed $(G_1\times G_2)'\subseteq G_1'\times G_2'$, and you'll also want to show the other direction. – 2017-01-03
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    In reference to the edit of your last comment, are you sure you know what $G'$ means? What are the elements of $G'$? What do they all look like? What can you say about their form? – 2017-01-03
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    @arctictern, I know as much as I wrote up there. I know what a subgroup is. I think I know what it means to be generated by something - it's been a while, and I'm a little rusty. But it's best to assume Jessy doesn't know what anything means. I thought the elements of $G^{\prime}$ were just all the commutators of the elements of $G$. – 2017-01-03
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    @arctictern now that I look more at what you wrote, it looks more like an internal direct product to me than an external one, which is what I'm going for here. – 2017-01-03
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    If you're working internally then just omit the $(\cdot,\cdot)$s in all of the $(a,b)$s and $(c,d)$s. – 2017-01-03
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    Not working internally. Working externally. – 2017-01-03
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    Oh. Then not sure where you're getting your impression from. Everything I wrote is about external direct products. – 2017-01-03
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    @arctictern wait a minute. I think we've had a similar conversation before. When you say "products" you mean under the group operation, and of course $G^{\prime}$ if it is generated by commutators of elements of $G$, it wouldn't just consist of those commutators. It needs to be closed under the group operation, so it would need to contain all the products of those elements under that group operation. Like I said, rusty. – 2017-01-03
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    So, I'm going to try to rework and repost. – 2017-01-03
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    @arctictern I just posted a new answer. – 2017-01-03