# Subgroups of direct products of free groups

I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/

He says that if $$G= F_{1} \times F_{2}$$ is a direct product of two free groups and $$H$$ is a subgroup of $$G$$, then it can be assumed that the projection maps $$\text{p}_{i} \colon H \rightarrow F_{i}$$ are surjective.

I do not understand why is this trivial. If $$\{f_{1},\cdots,f_{n}\}$$ is a basis of the free group $$F_{1}$$, why should we have elements of the form $$(f_{i},h_{i})$$ in $$H$$ for all $$i\in \{1,\cdots,n\}$$? He simply says that this follows because subgroups of free groups are free.

• Let $H_i\subset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $H\subset H_1\times H_2$ – Max Jan 3 at 17:37
• Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :) – Karen Jan 3 at 17:44
• Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica". – Shaun Jan 3 at 17:51