# Let $H_i$ be a subgroup of $G_i$ for $i=1,2,\dots,n.$ Prove that $H_1×\dots × H_n$ is a subgroup of $G_1 ×\dots × G_n.$

First I suuuuuck at proofs. I think I am on the right track but I need some fine tuning. Or if I am totally off let me know.

First we show that $$H_i$$ is nonempty.

Note that since $$H_i$$ is a subgroup of $$G_i, H_i$$ contains the identity element. So $$e_G \in H_1, e_G \in H_2,..., e_G \in H_n.$$ This means that $$e_G \in H_1 \times H_2 \times \cdots \times H_n.$$ Hence $$H_i$$ is non empty.

Next we show it is a subset. Let $$(x_1,x_2,...,x_n) \in H_1 \times H_2 \times...\times H_n.$$ This means $$x_1 \in H_1, x_2 \in H_2,...x_n \in H_n.$$ Since $$H_i \leq G_i, x_1 \in G_1, x_2 \in G_2,...x_n \in G_n,$$ which shows that $$(x_1,x_2,...,x_n) \in H_1 \times H_2 \times...\times H_n$$ and therefore $$\in G_1 \times G_2 \times...\times G_n.$$ Hence $$H_1 \times H_2 \times...\times H_n \subseteq G_1 \times G_2 \times...\times G_n.$$

Let $$x= (x_1,x_2,...,x_n)$$ and $$y= (y_1,y_2,...,y_n) \in H_1 \times H_2 \times...\times H_n.$$ Then $$xy^{-1}= (x_1,x_2,...,x_n)(y_1^{-1},y_2^{-1},...,y_n^{-1})= (x_1y_1^{-1},x_2y_2^{-1},...,x_ny_n^{-1}).$$ Then $$x_1y_1^{-1} \in H_1, x_2y_2^{-1} \in H_2,..., x_ny_n^{-1} \in H_n.$$ This means that $$(x_1y_1^{-1},x_2y_2^{-1},...,x_ny_n^{-1}) \in H_1 \times H_2 \times \cdots \times H_n$$ since $$H_i \leq G_i$$ and inverses are elements in the group.

Hence $$H_1 \times H_2 \times \cdots \times H_n$$ is a subgroup of $$G_1 \times G_2 \times \cdots \times G_n$$.

How about doing it for $$n=2$$, and then iterating?

Also, how about using the subgroup criterion? Namely, if $$h_1,h_2\in H\implies h_1h_2^{-1}\in H$$, then $$H\le G$$.

So take $$(a_1,b_1),(a_2,b_2)\in H_1×H_2$$. Just check that $$(a_1,b_1)(a_2,b_2)^{-1}=(a_1,b_1)(a_2^{-1},b_2^{-1})=(a_1a_2^{-1},b_1b_2^{-1})\in H_1×H_2$$, which it is since $$H_1,H_2$$ are subgroups.

Now, looking at what you did, I think you've got it. I see you also used the subgroup criterion. I guess my solution was largely hot air, because you already did it.

Taking one more look, let me fine tune a little bit. I don't think you need to separately show $$e$$ is in the subset. Because, it follows from $$xy^{-1}\in H$$, by setting $$y=x$$. Of course, you did note that $$H=H_1×\dots ×H_n$$ is non-empty already. But that does also follow from the fact that the $$H_i$$ are all non-empty.

On second thought, your way of doing it was quite right: you get $$e$$ pretty much immediately. Even if you haven't proved it is $$e$$ yet.

• Sometimes I like to talk in circles in my proofs or put unnecessary things in. My professor however grumbles when we don't specifically show its nonempty but I suppose there is different way to show it like you pointed out. Just want to make sure the way I showed it wasn't totally off. Thanks for your input. – PhysicsBish Mar 28 '20 at 4:31
• Sure. I'm guilty of rambling a bit here myself. I think you did a good job with it. – Chris Custer Mar 28 '20 at 4:39
• Awesome thanks again! – PhysicsBish Mar 28 '20 at 5:29